## An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 A° .

Q: An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 A° .

Solution: For an electron

$\large \frac{1}{2} m u^2 = e V$

(Where V is accelerating potential)

$\large \lambda = \frac{h}{m u}$

$\large \frac{1}{2} m (\frac{h}{m \lambda})^2 = e V$

$\large V = \frac{h^2}{2 m e \lambda^2}$

$\large V = \frac{(6.625 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (1.54 \times 10^{-10})^2}$

= 63.3 volt

## A compound of vanadium has a magnetic moment of 1.73 BM work out the electronic configuration of the vanadium in the compound.

Q: A compound of vanadium has a magnetic moment of 1.73 BM work out the electronic configuration of the vanadium in the compound.

Solution: Magnetic moment  $\large = \sqrt{n(n+2)}$

Where n is number of unpaired electrons

$\large 1.73 = \sqrt{n(n+2)}$

or,  (1.73)2 = n 2 + 2 n

(√3)2 = n 2 + 2 n

n 2 + 2 n -3 = 0

n 2 + 3 n – n -3 = 0

(n + 3)(n-1) = 0

⇒  n = 1 , since n ≠ -3

Vanadium atom must have the unpaired electron and thus its configuration is:

23V4+: 1s2 2s2 2p6 3s2 3p6 3d1

## Given below are the sets of quantum numbers for given orbitals. Name these orbitals.

Q: Given below are the sets of quantum numbers for given orbitals. Name these orbitals.

(i) n = 2, l = 1, m = –1

(ii) n = 4, l = 2, m = 0

(iii) n = 3, l = 1, m = ± 1

(iv) n = 4, l = 0, m = 0

(v) n = 3, l = 2, m = ± 2

Solution: (i) 2px or 2py

(ii) 4dz2

(iii) 3px or 3py

(iv) 4s

(v) 3dx2y2 or 3 dxy

## Find out the angular momentum of an electron in (i) 4s orbital (ii) 3p orbital (iii) 4th orbital

Q: Problem : Find out the angular momentum of an electron in

(i) 4s orbital

(ii) 3p orbital

(iii) 4th orbital

Solution: Angular momentum in an orbital $\large = \frac{h}{2 \pi} \sqrt{l (l+1)}$

(i) l = 0 for 4s orbital, hence orbital angular momentum = 0

(ii) l = 2 for 3p orbital

Angular momentum $\large = \frac{h}{2 \pi} \sqrt{1 (1 + 1)} = \frac{h}{\sqrt{2}\pi}$

(iii) Angular momentum in an orbit $\large = \frac{n h}{2 \pi} = \frac{4 h}{2 \pi} = \frac{2h}{\pi}$

## Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits a photon of wavelength 121.6 nm. Which transition leads to this wavelength? How fast were the hydrogen atoms travelling before collision ?

Q: Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits a photon of wavelength 121.6 nm. Which transition leads to this wavelength? How fast were the hydrogen atoms travelling before collision ?

Solution: Wavelength is emitted in UV region and thus n1 = 1

For H atom , $\large \frac{1}{\lambda} = R_h (\frac{1}{1^2} – \frac{1}{n^2})$

$\large \frac{1}{121.6 \times 10^{-9}} = 1.097 \times 10^7 (\frac{1}{1^2} – \frac{1}{n^2})$

Hence , n = 2

Also the energy released is due to collision and all the kinetic energy is released in form of photon.

$\large \frac{1}{2} m v^2 = \frac{h c}{\lambda}$

$\large \frac{1}{2} \times 1.67 \times 10^{-27} \times v^2 = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{121.6 \times 10^{-9}}$

v = 4.43 × 104 m/sec

## Iodine molecule dissociates into atoms after absorbing light of 4500 A° . If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms.

Q: Iodine molecule dissociates into atoms after absorbing light of 4500 A° . If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms. (Bond energy of I2 = 240 kJ (mol).

Solution: Energy given to iodine molecule

$\large \frac{h c}{\lambda} = \frac{6.625 \times 10^{-34}\times 3 \times 10^8}{4500 \times 10^{-10}}$

= 4.417 × 10–19J

Also energy used for breaking up I2 molecule

$\large = \frac{240 \times 10^5}{6.023 \times 10^{23}}$

= 3.984 × 10–19J

Energy used in imparting kinetic to two atoms

= (4.417 – 3.984) × 10–19 J

KE of iodine atom $\large = \frac{4.417 – 3.984}{2}\times 10^{-19}$

= 0.216 × 10–19 J

## When certain metal was irradiated with light frequency 1.6 × 10^16 Hz the photo electrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 1.0 × 10^16 Hz. Calculate threshold frequency for the metal.

Q: When certain metal was irradiated with light frequency 1.6 × 1016 Hz the photo electrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 1.0 × 1016 Hz. Calculate threshold frequency for the metal.

Sol: In first case ,

h ν1 = h ν0<>/sub + KE1

KE1 = h ν1 – h ν0

KE1 = h ( ν1 – ν0 ) ….(i)

Similarily ,

KE2 = h ( ν2 – ν0 )

KE1/2 = h ( ν2 – ν0 ) ….(ii)

On dividing (i) by (ii)

$\large 2 = \frac{\nu_1 – \nu_0}{\nu_2 – \nu_0}$

$\large 2 = \frac{1.6 \times 10^{16} – \nu_0}{1 \times 10^{16} – \nu_0}$

ν0 = 4 × 1015 Hz