An electron is accelerated through a potential difference of V volt. It has a wavelength λ associated with it…..

Q. An electron is accelerated through a potential difference of V volt. It has a wavelength λ associated with it. Through what potential difference an electron must be accelerated so that its de Broglie wavelength is the same as that of a proton? Take mass of proton to be 1837 times larger than the mass of electron.

(a) V volt

(b) 1837V volt

(c) V/1837 volt

(d) √(1837)volt

Ans: (d)

Sol: λ = h/√(2meV)

λ’ = h/√(2m’eV’)

Since , λ = λ’

h/√(2meV) = h/√(2m’eV’)

m V =m’V’

m V = m×1837×V’

V’ = V/1837 volt

 

 

When a metallic surface is illuminated by a light of frequency 8×10^14 Hz, photoelectron of maximum energy …..

Q. When a metallic surface is illuminated by a light of frequency 8×1014 Hz, photoelectron of maximum energy 0.5eV is emitted. When the same surface is illuminated by light of frequency 12 × 1014 Hz, photoelectron of maximum energy 2eV is emitted. The work function is

(a) 0.5 eV

(b) 2.85 eV

(c) 2.5 eV

(d) 3.5 eV

Ans: (c)

Sol: In first case ,

h(8×1014) = φ+0.5    ….(i)

In 2nd case ,

h(12 × 1014) = φ+2  ….(ii)

Dividing (ii) by (i)

h(12 × 1014) /h(8×1014)= (φ+2)/(φ+0.5)

3/2 =  (φ+2)/(φ+0.5)

2φ + 4 = 3φ + 1.5

φ = 2.5 eV

A modern 200W sodium street lamp emits yellow light of wavelength 0.6mm. Assuming it to be 25% efficient …..

Q. A modern 200W sodium street lamp emits yellow light of wavelength 0.6mm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

(a) 62 × 1020

(b) 3 × 1019

(c) 1.5 × 1020

(d) 6 × 1018

Ans: (c)

Sol: Energy of photon , E= hc/λ

E = 6.6×10-34 × 3× 108/(0.6× 10-6)

E = 3.3× 10-19 J

Efficiency , η = P0/PI

25/100 = P0/200

P0 = 50 W

Number of photons emitted per sec

N = P0/E = 50/3.3× 10-19 = 1.5× 1020