Q: Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom?

For Balmer Series: Shortest wavelength, E o 4 = h c λ s Longest wavelength, E o 4 − E 0 9 = h c λ l = 5 E o 36 ∴ λ l λ s = 9 5 Range of energy level in Balmer series, 5 E 0 36 , E 0 4 Range of energy level in Paschen series, 7 E 0 144 , E 0 9 Hence, no overlap. For Lyman series: For shortest wavelength, E 0 = h c λ 0 For general wavelength, E 0 - E 0 n 2 = h c λ ⇒ 1 1 − 1 n 2 = λ λ 0 ∴

Q: A particle of mass m moves in circular orbits with potential energy U r = F r , where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model.

U = F · r F c = - d U d r = - F m v 2 r = F m u r = n h 2 π ⇒ m u r 2 m u 4 r = n 2 h 2 4 π 2 · F ⇒ m r 3 = n 2 h 2 4 π 2 F ⇒ r = n 2 h 2 4 π 2 m F 1 / 3 ⇒ r ∝ n 2 / 3 E = U + K = F · r + 1 / 2 m v 2 = F · r + 1 / 2 F · r E = 3 / 2 F · r Put value of r , E = 3 2 F r n 2 h 2 4 π 2 m F 1 / 3

Question 1

Consider a hydrogen atom with v_k, r_k, and K_k denoting the velocity, orbital radius and kinetic energy of the electron in the k^ th orbit, respectively. The electron undergoes a transition from the

Accepted Answer

From Bohr's quantization condition, the angular momentum of an electron in the k^ th orbit is given by: m v_k r_k = k h 2 The kinetic energy of the electron in the k^ th orbit is: K_k = 1 2 m v_k^2 = 1 2 (m v_k r_k) v_k r_k Substituting the value of angular momentum: K_k = 1 2 ( k h 2 ) v_k r_k = k h v_k 4 r_k The magnitude of change in kinetic energy for a transition from the n^ th orbit to the 1^ st orbit is: | K| = |K_n - K_1| = h 4 | n v_n r_n - v_1 r_1 | This makes statement (A) correct. Si

Question 2

Consider an electron in the n=3 orbit of a hydrogen-like atom with atomic number Z. At absolute temperature T, a neutron having thermal energy k_ B T has the same de Broglie wavelength as that of this

Accepted Answer

mv ^2 r = KZe ^2 r ^2 mv ^2 r = 1 4 _0 Ze ^2 mvr = nh 2 (1)/(2) gives v = Ze ^2 4 _0 nh 2 = Ze ^2 2 _0 nh aligned & h m v = h 2 m_N K_B T \ & T= m^2 Z^2 e^4 8 _0^2 n^2 h^2 m_N K_B \ & n=3 T= m^2 Z^2 e^4 72 _0^2 h^2 m_N K_B \ & (1) (2)^2 1 m r = Z e^2 4 _0 n^2 h^2 4 ^2 aligned aligned & r = n ^2 ~h ^2 _0 Ze ^2 ~m a _0= h ^2 _0 e ^2 ~m \ & a _0^2= h ^4 _0^2 ^2 e ^4 ~m ^2 \ & Ta _0^2= m ^2 Z ^2 e ^4 72 _0 ~h ^2 ~m _ N k _ B h ^4 _0^2 ^2 e ^4 ~m ^2 \ & ~T = h ^2 Z ^2 72 ^2 a _0^2 ~m _ N k _ B

Question 3

List-I shows various functional dependencies of energy (E) on the atomic number (Z). Energies associated with certain phenomena are given in List-II. the option that describes the correct match betwee

Accepted Answer

(P) Energy of H -like atom is aligned & E =-13.6 Z ^2 n ^2 So \ & E Z ^2 \ & P (5) aligned (Q) Energy of characteristic X-ray by moseley's correction aligned & E=-13.6(Z-1)^2 [ 1 n_1^2 - 1 n_2^2 ] So \ & E (Z-1)^2 \ & Q (1) aligned (R) Electrostatics binding energy is proportional to Z(Z-1) R (2) (S) For stable nuclei with mass no. in range 30 to 170 . Binding energy per nucleon is constant \& graph is straight line S (4) Ans. (C) is correct

Question 4

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v_1 and ejects the electron with a kinetic energy of 10 eV . The electron then combines with a positron at rest to

Accepted Answer

h v_1=13.6+10=23.6 eV ....(1) Energy of positronium in ground state aligned & =-13.6 m ( z n )^2 eV \ & =-13.6 1 2 eV =-6.8 eV aligned So to make positronium 6.8 eV must release \& 5 eV is the KE of COM . So total energy of photon released (h v_2 ) will be : h v_2=(10-5)+6.8=11.8 eV ....(2) Difference in energy =23.6-11.8=11.8 eV

Question 5

A particle of mass m is moving in a circular orbit under the influence of the central force F(r)=-k r, corresponding to the potential energy V(r)=k r^2 / 2, where k is a positive force constant and r

Accepted Answer

The central force will provide necessary centripetal force kr = mv ^2 r or, kr ^2= mv ^2 ...(1) By quantisation rule n = mvr or, n r = mv ...(2) aligned & (1) (2)^2 kr ^2 n ^2 ^2 r ^2 = mv ^2 ~m ^2 v ^2 \ & k n ^2 ^2 r ^4= 1 ~m \ & r = ( n ^2 ^2 ~km )^ 1 4 r ^2= n mk aligned aligned & (2) Using (1), K n mk = mv ^2 \ & v ^2= n k m ^3 aligned (3) L mr ^2 = mvr mr ^2 = v r = k m from (1) aligned & (4) E = 1 2 mv ^2+ 1 2 kr ^2= n 2 k m + 1 2 k n mk \ & E = n k m aligned

Question 6

Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom?

Accepted Answer

For Balmer Series: Shortest wavelength,  E o 4 = h c &#955; s Longest wavelength,  E o 4 &#8722; E 0 9 = h c &#955; l = 5 E o 36 &#8756; &#160; &#955; l &#955; s = 9 5 Range of energy level in Balmer series, 5 E 0 36 , &#160; E 0 4 Range of energy level in Paschen series, 7 E 0 144 , &#160; E 0 9 Hence, no overlap. For Lyman series: For shortest wavelength, E 0 = h c &#955; 0 For general wavelength, E 0 - E 0 n 2 = h c &#955; &#8658; 1 1 &#8722; 1 n 2 = &#955; &#955; 0 &#8756;

Question 7

A particle of mass m moves in circular orbits with potential energy U r = F r , where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model.

Accepted Answer

U = F &#183; r F c = - d U d r = - F m v 2 r = F m u r = n h 2 &#960; &#8658; m u r 2 m u 4 r = n 2 h 2 4 &#960; 2 &#183; F &#8658; m r 3 = n 2 h 2 4 &#960; 2 F &#8658; r = n 2 h 2 4 &#960; 2 m F 1 / 3 &#8658; r &#8733; n 2 / 3 E = U + K = F &#183; r + 1 / 2 &#160; m v 2 = F &#183; r + 1 / 2 F &#183; r E = 3 / 2 F &#183; r Put value of  r , E = 3 2 F r n 2 h 2 4 &#960; 2 m F 1 / 3

Atomic Physics — Previous Year Questions & Solutions