Q: The stopping potential (V_0 ) versus frequency (v) of a graph for photoelectric effect in a metal. From the graph, the planck's constant (h) is.

aligned e V_0=h (v-v_0 ) & =h v-h v_0 \\ V_0 & = ( h e ) v- ( h e ) v_0 (i) aligned From Eq. (i), it follows that the slope of the graph is h e . Therefore, aligned h=e slope & =1.6 10^ -19 165-0 (8-4) 10^ 15 \\ & = 1.6 10^ -19 16.5 4 10^ 15 =6.6 10^ -34 ~J - s . aligned

Question 1

Which of the following transitions of He ^ + ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom?

Accepted Answer

Let the transition of electron in hydrogen atom (Z=1) gives wavelength _1 for transition from n_2 to n_1, then 1 =R Z^2 ( 1 m^2 - 1 n^2 ) Here, = _1, Z=1, m=n_1, n=n_2 1 _1 =R(1)^2 ( 1 n_1^2 - 1 n_2^2 )=R ( 1 n_1^2 - 1 n_2^2 ) Now, if electron makes transition in He ^ + ion from n_4 to n_3 and gives wavelength _2, then aligned 1 _2 & =R(2)^2 ( 1 n_3^2 - 1 n_4^4 ) \ & =R [ 1 ( n_3 2 )^2 - 1 ( n_4 2 )^2 ] aligned Given that, _1= _2 R ( 1 n_1^2 - 1 n_2^2 )=R [ 1 ( n_3 2 )^2 - 1 ( n_4 2 )^2 ] On co

Question 2

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is:

Accepted Answer

Wavelength of H -atom is 1 = Rz ^2 ( 1 n _1^2 - 1 n _2^2 ) Shortest wavelength for Balmer series: 1 _B = Rz ^2 ( 1 2^2 - 1 ) Shortest wavelength for Lyman series: 1 _ L = Rz ^2 ( 1 1^2 - 1 ) ....(ii) Dividing eq. (i) and (ii), _ A : _1=4: 1

Question 3

The minimum excitation energy of an electron revolving in the first orbit of hydrogen is

Accepted Answer

Energy of electron in n ^ th orbit E_n= -13.6 e V n^2 Excitation energy of electron from n=1 to n=2 E=-13.6 [ 1 2^2 - 1 1 ]=-13.6 [ 1 4 - 1 1 ]=10.2 eV

Question 4

The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

Accepted Answer

From the Bohr's quantization rule Angular momentum L = nh 2 , L _1= 1 . h 2 = L ( n=1 for first orbit) in ( n=2 for second orbit) L_2= 2 h 2 =2 L Hence, change in angular momentum L _2- L _1=2 ~L - L = L

Question 5

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

Accepted Answer

If we assume electron in hydrogen atom takes energy 12.09 eV from the incoming radiation, the maximum excited state of electron will be n=3. So, number of spectral lines is 3(3-1) 2 =3. Here we assume some part of energy 12.5 eV -12.09 eV =0.41 eV get lost due to collision.

Question 6

The stopping potential (V_0 ) versus frequency (v) of a graph for photoelectric effect in a metal. From the graph, the planck's constant (h) is.

Accepted Answer

aligned e V_0=h (v-v_0 ) & =h v-h v_0 \ V_0 & = ( h e ) v- ( h e ) v_0 (i) aligned From Eq. (i), it follows that the slope of the graph is h e . Therefore, aligned h=e slope & =1.6 10^ -19 165-0 (8-4) 10^ 15 \ & = 1.6 10^ -19 16.5 4 10^ 15 =6.6 10^ -34 ~J - s . aligned

Question 7

The significant result deduced from the Rutherford's scattering experiment is that

Accepted Answer

The significant result deduced from the Rutherford's scatter ing is that whole of the positive charge is concentrated at the centre of atom i.e. nucleus.

Atomic Physics — Previous Year Questions & Solutions