Question 1
What is the angular momentum of an electron in the third orbit of an atom?
(a) $3.15 \times 10^{-34}$ Js
(b) $3.15 \times 10^{-30}$ Js
(c) $3.15 \times 10^{-31}$ Js
(d) $3.15 \times 10^{-33}$ Js Solution
Angular momentum is given by
$L= \frac {nh}{2\pi} = \frac {3 \times 6.6 \times 10^{-34} \times 7}{2 \times 22}=3.15 \times 10^{-34}$ Js
So (a) is correct
Question 2
Which series of the hydrogen spectrum has wavelength in the visible range
(a) Lyman.
(b) Balmer.
(c) Paschen.
(d) Bracket. Solution
Balmer
Hence (b) is the correct option
Question 3
The value of fine structure constant is ?
(a) $\frac {1}{136}$
(b) $\frac {1}{130}$
(c) $\frac {1}{138}$
(d) $\frac {1}{137}$ Solution
(d)
Question 4
If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation is ?
(a) 5R/36
(b) 36/5R
(c) R/6
(d) 5/R Solution
$\bar{\nu} = \frac {1}{\lambda}= R (\frac {1}{2^2} - \frac {1}{3^2})$
or $\lambda= \frac {36}{5R}$
(b) is correct option
Question 5
if an atom moves from 2E energy level to E energy level,the wavelength $\lambda$ is emitted. If the transition takes place from 4E/3 energy level to E energy level, the wavelength emitted will be
(A) $\frac {\lambda}{3}$
(B) $3 \lambda$
(C) $\frac {3\lambda}{4}$
(D) $\frac {4\lambda}{3}$ Solution
case -1
$2 E - E = E= \frac {hc}{\lambda}$
Case -2
$\frac {4E}{3} - E = \frac {hc}{\lambda _1}$
or
$\lambda _1 =3 \lambda$
Hence (b) is correct
Question 6
Maximum frequency of the emission is obtained for the transition
(a) n=2 to n=1
(b) n=6 to n=2
(c) n=1 to n=2
(d) n=2 to n=6 Solution
(a) as the energy difference is maximum
Question 7
The radius of an electron orbit in a hydrogen atom is of the order of
(a) $10^{-8}$ m.
(b) $10^{-9}$ m.
(c) $10^{-11}$ m.
(d) $10^{-13}$ m. Solution
Hence (c) is correct
Question 8
If an electron jumps from Ist orbit to 4th Orbit then it will
(a) not lose energy.
(b) absorb energy.
(c) release energy.
(d) increases and decreases periodically Solution
(b)
Question 9
The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a He+ ion in the first excited state will be
(a)-13.6 eV
(b)-6.8 eV
(c)-54.4.6 eV
(d)-27.2 eV Solution
for firt excited state of He+ ion
n=2, Z=2
$E= = 13.6 \frac {Z^2}{n^2} = -13.6$ eV
Hence (a) is the correct choice
Question 10
Energy of an electron in the second orbit of hydrogen atom is E and energy of electron in 3rd orbit($E_3$) of He+ will be
(a) $E_3 = \frac {16E}{3}$
(b) $E_3 = \frac {16E}{9}$
(c) $E_3 = \frac {4E}{9}$
(d) $E_3 = \frac {4E}{3}$ Solution
$E= = E_0 \frac {Z^2}{n^2}$
For H atom, Z=1, n=2
$E= = E_0 \frac {1^2}{2^2}= \frac {E_0}{4}$
For He+ atom, Z=2, n=3
$E_3= = E_0 \frac {2^2}{3^2}= \frac {4E_0}{9}$
Therefore
$E_3= \frac {16}{9} E$
Hence (b) is correct
Question 11
The ratio of minimum to maximum wavelength in balmer series is
(a) 1 :4.
(b) 5 : 9
(c) 3 :4
(d) 5 : 36 Solution
balmer series of hydrogen spectrum is given
$\frac {1}{\lambda} = R (\frac {1}{2^2} - \frac {1}{n^2}) \; m^{-1}$ where n= 3,4,5
For max Wavelength
$\frac {1}{\lambda _{max}} = R (\frac {1}{2^2} - \frac {1}{3^2})= \frac {5R}{30}$
For min Wavelength
$\frac {1}{\lambda _{min}} = R (\frac {1}{2^2} - \frac {1}{\infty ^2})= \frac {R}{4}$
Therefore $ \frac {\lambda _{min}}{\lambda _{max}} \frac {5}{9}$
hence (b) is correct
Question 12
The short wavelength limits of the lyman, Paschen and Balmer series in the hydrogen spectrum are a , b and c respectively . Arrange these wavelength in the increasing order?
(a)a < b < c
(b)c < b < a
(c)a < c < b
(d)b < c < a Solution
(c) is correct
Question 13
With increasing quantum number, the energy difference between adjacent level
(a) increases
(b) decreases
(c) same
(d) increase and then decreases Solution
Question 15
(a) a -> p, b -> q , c -> r.
(b) a -> r, b -> q , c -> p.
(c) a -> r, b -> p , c -> q.
(d) a -> p, b -> r , c -> q. Solution
(c)
Question 16
The ground state energy of the hydrogen atom is $E_0$. The kinetic energy of the electron in the 3rd excitel level
(a) $-\frac {E_0}{16}$
(b) $-\frac {E_0}{9}$
(c) $\frac {E_0}{16}$
(d) $\frac {E_0}{9}$ Solution
$E= \frac {E_0}{n^2}$
For third excited level, n=4
$E_4= \frac {E_0}{16}$
Kinetic energy = $-E_4= -\frac {E_0}{16}$
hence (a) is correct
Question 17
The ground state energy of the hydrogen atom is $E_0$. The Potential energy of the electron in the 2nd excitel level
(a) $\frac {E_0}{8}$
(b) $\frac {E_0}{9}$
(c) $\frac {E_0}{16}$
(d) $\frac {2E_0}{9}$ Solution
$E= \frac {E_0}{n^2}$
For third excited level, n=3
$E_3= \frac {E_0}{9}$
Potential energy = $2E_3= \frac {2E_0}{9}$
hence (d) is correct
Question 18
The electron in hydrogen atom is initially in the third excited state.What is the maximum number of lines of spectrum lines which can be emitted whne it finally moves to the ground state
(a) 3
(b) 4
(c) 5
(d)6 Solution
Here n=4
Number of spectrum lines is given by = $\frac {n(n-1)}{2} = 6$
hence (d) is correct
Question 19
A set of atoms in an excited state decays.
(a) in general to any of the states with lower energy.
(b) into a lower state only when excited by an external electric field.
(c) all together simultaneously into a lower state.
(d) to emit photons only when they collide Solution
(a)
Question 20
A hydrogen atom and a Li++ ion are both in the second excited state. If $I_H$ and $I_{Li}$ are their respective electronic angular momentum and $E_H$ and $E_{Li}$ their respective energies,then
(a) $I_H=I_{Li}$ , $|E_{Li}| > |E_H|$
(b) $I_H < I_{Li}$ , $|E_{Li}| < |E_H|$
(c) $I_H > I_{Li}$ , $|E_{Li}| > |E_H|$
(d) $I_H=I_{Li}$ , $|E_{Li}| < |E_H|$ Solution
(a)
Text Based Questions(long)
Question 1
Define the distance of closest approach. An alpha-particle of kinetic energy 'K' is bombarded on a thin gold foil. The distance of the closest approach is 'r'. What will be the distance of closest approach for an alpha-particle of double the kinetic energy?
Question 2
Using Bohr's postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram
showing how the line spectra corresponding to Ballmer series occur due to transition between energy levels.
Question 3
(i)State Bohr's postulates to define stable orbits in hydrogen atom. How does de -broglie hypothesis explain the stability of these orbits?
(ii) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n=4 level . Estimate the frequency of the photon Solution
Question 4
Derive the Bohr's Quantisation condition for angular momentum of the orbitting of electron in hydrogen atom using de Broglie hypothesis.
Question 5
Using Bohr's postulates of the atomic model, derive the expression for the radius of the nth electron orbit.Also obtain the expression for Bohr's Radius. Show graphically the variation of the radius of the orbit with the principle quantum number
Short Answer type
Question 1
Calculate the ratio of the frequencies of the radiation emitted due to transition of the electron in a hydrogen atom from its (i) second permitted energy level to the first level and (ii) highest permitted energy level to the second permitted level Solution
Energy of an electron in nth permitted energy level is given by
$E_{nth}=- \frac {13.6}{n^2}$ eV.
(i) Energy of photon produced due to transition of an electron from second permitted energy level to the first level
$E=E_2-E_1$
$E=-\frac {13.6}{(2)^2} - \frac {-13.6}{1^2}=10.2$eV
(ii) Energy of photon produced due to transition of an electron from highest permitted energy level to the second permitted energy level.
$E^{'}=E_{\infty}-E_2$
$E^{'}=- \frac {13.6}{(\infty)^2} - \frac {-13.6}{2^2}$
=3.4 eV
Ratio of (i) and (ii)
$\frac {E}{E^{'}} =\frac {10.2}{3.4}=3$
Question 2
Calculate the shortest wavelength of light emitted in the Paschen series of hydrogen spectrum. Which part of the electromagnetic spectrum , does it belong?br>
Explain the above observations on the basis of Einstein's photoelectric equation.? Solution
Paschen series of hydrogen spectrum is given
$\bar{\nu} = 1.1 \times 10^{7} (\frac {1}{3^2} - \frac {1}{n^2}) \; m^{-1}$ where n= 4,5,6
$\frac {1}{\lambda} = 1.1 \times 10^{7} (\frac {1}{3^2} - \frac {1}{n^2}) \; m^{-1}$
For Shortest wavelength, $n = \infty$
$\frac {1}{\lambda} = 1.1 \times 10^{7} (\frac {1}{3^2} - \frac {1}{\infty ^2}) $
$\lambda = 8199 A^o$
The series lies in infrared region of the EM spectrum
Question 3
A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelength and the corresponding series of the lines emmitted Solution
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e., -1.1 eV.
Energy is related to orbit level (n) is given as
$E_{nth}=- \frac {13.6}{n^2}$ eV.
For n=3
$E_{3}=- \frac {13.6}{3^2}= -1.5$ eV.
For n=4
$E_{4}=- \frac {13.6}{4^2}= -.85$ eV.
Since -1.1 eV lies between $E_3$ and $E_4$ ,So it is clear, it can go upto level 3 when 12.5 eV energy is supplied
Now from n= 3, Hydrogen can go from
a. n=3 to n=1
b. n=3 to n=2 and then n=2 to n=1
For n=3 to n=2, it will be Balmer series
$\frac {1}{\lambda} = R (\frac {1}{2^2} - \frac {1}{3^2}) $
Now $R= 1.7 \times 10^7$
Hence Wavelength will be given as 656.33 nm
For n=3 to n=1, it will be Lyman series
$\frac {1}{\lambda} = R (\frac {1}{1^2} - \frac {1}{3^2}) $
Now $R= 1.7 \times 10^7$
Hence Wavelength will be given as 102.55 nm
For n=2 to n=1, it will be Lyman series
$\frac {1}{\lambda} = R (\frac {1}{1^2} - \frac {1}{2^2}) $
Now $R= 1.7 \times 10^7$
Hence Wavelength will be given as 121.5 nm
So in Summary, It will emit radiation of wavelength 102.55 nm and 121.5 nm in Lyman series and 656.33 nm in Balmer series
Question 4
Find the relation between the three wavelength $\lambda _1$,$\lambda _2$ and $\lambda _3$ from energy level diagram shown below? Solution
Question 6
Draw the energy-level diagram of hydrogen atom. This atom in its ground state is excited by absorption of radiation of wavelength 970 $A^o$. How many different kinds of lines are possible in the emitting spectrum?. calculate the longest wavelength among them.Take ionization energy of hydrogen atom as 13.6 eV, h=$6.6 \times 10^{-34}$ Js. c = $3 \times 10^8$ m/s ,1 eV = $1.6 \times 10^{-19}$ J Solution
The energy given is
$E= \frac {hc}{\lambda}$
Subsitituting the values
$R= 12.75$ eV
The energy of hydrogen in ground state is -13.6 eV. Hence the energy of the excited atom will be
=-13.6 + 12.75 = -.85 eV
Energy is related to orbit level (n) is given as
$E_{nth}=- \frac {13.6}{n^2}$ eV.
Therefore
$-.85=- \frac {13.6}{n^2}$
or n=4
Now from this energy level, following transition are possible
4 -> 3, 4 -> 2, 4 -> 1 , 3 ->2 , 2 -> 1,3 ->1
So total 6 lines are possible in the emitting spectrum
Now the longest wavelength will be emitted when the transition happens from n=4 to n=3
Energy in the third orbit is given by
$E_{3}=- \frac {13.6}{3^2} = -1.51$ eV.
Now energy emitted when this transition happens
$\Delta =E_4 - E_3 =-.85 - (-1.51) =.66 eV$
Now Wavelength will be given as
$\Delta E= \frac {hc}{\lambda}$
or $\lambda = 18750 A^0$
Numerical Questions
Question 1
(i)The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm?
(ii)which transition correspond to emission of radiation of maximum wavelength?
Solution
(i) Energy transitions for A,B,C, and D are:
A = 2 eV
B = 4.5 eV
C = 2.5 eV
D = 8 eV
Now
$E= \frac {h c}{\lambda}$
$\lambda = \frac { hc}{E}$
Now
$h = 6.63 \times 10^{-34}$ Js
$c = 3 \times 10^8$ m/s
Lets calculate the Wavelength for each of the above energy transition
For A
$\lambda = \frac { hc}{E_A} = 621.1$ nm
For B, we have
$\lambda = \frac { hc}{E_B} = 276$ nm
For C, we have
$\lambda = \frac { hc}{E_C} = 496.9$ nm
For D, we have
$\lambda = \frac { hc}{E_D} = 155.2$ nm
So energy transition B will show photon of wavelength 275 nm
(ii) Energy transition B has radiation of maximum wavelength 621 nm
Question 2
In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z=80, when an alpha-particle of 8MeV energy impings on it before it comes momentarily to rest and reverses its direction.
How will the distance of closest approach be affected when the kinetic energy of the a- particle is doubled ? Solution
Z=80,KE=8MeV
At closest approach
Potential Energy = Kinetic energy
$\frac {k Ze^2}{d} = 8 \times 1.6 \times 10^{-13}$
or d = 128.8 fm
We can see that d is inversely proportional to the Kinetic energy,So if kinetic energy of the alpha- particle is doubled, distance of closest approach is halved
Question 3
The energy of the electron, the hydrogen atom, is known to be expressible in the form
$E_{nth}=- \frac {13.6}{n^2}$ eV n=1,2,3...
Use this expression to show that the
(i) electron in the hydrogen atom can not have an energy of -6.8 eV.
(ii) spacing between the lines (consecutive energy levels) within the given set of the observed hydrogen atom spectrum decreases as n increases. Solution
$E_{nth}=- \frac {13.6}{n^2}$ eV n=1,2,3...
$E_1= - \frac {13.6}{1^2} = -13.6$ eV
$E_2= - \frac {13.6}{2^2} = -3.4$ eV , $E_2 -E_1=10.2$ eV
$E_3= - \frac {13.6}{3^2} = -1.51$ eV , $E_3 - E_2 = 1.89$ eV
$E_4= - \frac {13.6}{4^2} = -.85$ eV, $E_4 - E_3 = .66$ eV
$E_5= - \frac {13.6}{5^2} = -.54$ eV, $E_5 - E_4 = .31$ eV
(i)Clearly an electron in the hydrogen atom cannot have an energy of -6.8 eV
(ii) We can also clearly observed that spacing between the lines (consecutive energy levels) within the given set of the observed hydrogen atom spectrum decreases as n increases
Question 4
Photons, with a continuous range of frequencies, are made to pass through a simple of rarefied hydrogen. The transitions, shown in Fig. indicate three of the spectral absorption lines in the continuous spectrum.
(i) Identify the spectral series, of the hydrogen emission spectrum, to which each of these three lines correspond.
(ii) Which of these lines corresponds to the absorption of radiation of maximum wavelength? Solution
(i) Line I -> Lyman Series
Line II -> Balmer series
Line III -> Paschen series
(ii) Line III corresponds to absorption of the photon of minimum energy and hence maximum wavelength
Question 5
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Given $h=6.6 \times 10^{-34} $ Js,$c=3 \times 10^{8} $ m/s,$e=1.6 \times 10^{-19}$ C Solution
For Lyman series
$\frac {1}{\lambda} = R (\frac {1}{1^2} - \frac {1}{n^2}) $ n=2,3,4
Here $R=1.097 \times 10^7$ /m
substiuting these values and arranging the above equation
$\lambda= \frac {913.4 n^2}{n^2 -1} \times 10^7$
or
$\lambda= \frac {913.4 n^2}{n^2 -1} \; A^o $
Now Substituting values for n=2,3,4,5 , we get the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum
we get
$\lambda _{21} = 1218 \; A^o $
$\lambda _{31} = 1028 \; A^o $
$\lambda _{41} = 974.3 \; A^o $
and $\lambda _{51} = 951.4 \; A^o $
Question 6
A particle of charge equal to that of electron and mass 208 times the mass of electron ($\mu$ meson)moves in a circular orbit around a nucleus of charge +3e (assume mass of nucleus to be infinite). Assuming that the Bohr atom model is applicable to this system
(a) derive an expression for the radius of the nth Bohr orbit
(b) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom
(c) find the wavelength of the radiation emitted when -meson jumps from the third orbit to the first orbit (Rydberg constant = 1.097 x 107 m-1)
Ans
(a)$r=\frac {n^2 h^2 \epsilon _0}{624 \pi m e^2}$
(b)n=25
(c)$5.478 \times 10^{-11}$ m
Question 6
Suppose that the potential energy between an electron and a proton at a distance r is given by $-\frac {ke^2}{3r^3}$ . Use Bohr theory to obtain energy level of such a hypothetical hydrogen atom.
