DUAL NATURE OF MATTER AND RADIATION

11 )

Photocell cell is enlightened by small bright source 1 m away. If the same light source is placed (1/2) m away, number of electrons emitted by cathode will be……..

increases twice

decreases twice

increases 4 times

decreases 4 times

View Solution

Answer :

Intensity of light radiation = I ∝ (1 / r²)

distance r is changed by (1/2) hence intensity will increase by factor of (2)²

i.e. 4.

Hence rate of photoelectron emission from photo cathode also increases

by q factor of 4.

12 )

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be………..

√2

(1 / √2)

(1/2)

View Solution

Answer :

KE is made double, E₂ = 2E₁

λ = [h / √{2mE}]

i.e. λ ∝ {1 / √E}

∴ (λ₂ / λ₁) = √(E₁ / E₂)

∴ (λ₂ / λ₁) = √(1/2)

λ₂ = (1 / √2) λ₁

13 )

Energy corresponding to threshed frequency of metal is 6.2 eV. If stopping potential corresponding to radiation incident on surface is 5 V, incident radiation will be in the…………..…region.

X-ray

Ultraviolet

infrared

Visible

View Solution

Answer :

given: E = hV_o = 6.2 eV

V_s = 5V

using Einstein's photoelectric equation,

hƲ = hƲ_o + eV_s

hƲ = 6.2 eV + 5eV = 11.2 eV

∴ hƲ = 11.2 eV

∴ (hc / λ) = 11.2 eV

∴ λ = {hc / (11.2 eV)}

∴ λ [{6.6 × 10^–34 × 3 × 10⁸} / {11.2 × 1.6 × 10^–19}] = 1.1 × 10^–7 m = 1100Å

it corresponds to ultraviolet region.

14 )

At 10° C temperature, de-Broglie wave length of atom is 0.4Å .If temperature of atom is increased by 30° C, what will be change in de-Broglie wavelength of atom?

Select correct option from the following options

(A) decreases 10^–2 Å (B) decreases 2 × 10^–2 Å

View Solution

Answer :

T₁ = 10° C = 10 + 273 = 283K

T₂ = 10° + 30° = 40° C = 40 + 273 = 313K

λ₁ = 0.4Å = 0.4 × 10^–10m

λ = [h / {√2mE}]

i.e. λ ∝ (1 / √E) (1)

also E = (3/2) KT i.e. E ∝ T

hence equation (1) is

λ ∝ (1 / √T)

∴ (λ₂ / λ₁) = √(T₁ / T₂)

∴ λ₂ = λ₁ √(T₁ / T₂)

∴ λ₂ = 0.4 × 10^–10 √{(283) / (313)}

λ₂ = 0.38 × 10^–10 m

∴ λ₂ = 0.38 Å

Change in wavelength = λ₂ – λ₁

= (0.38 – 0.4)Å

= – 0.02Å

= 2 × 10^–2 Å decrement

15 )

Wavelength of incident radiation on photo sensitive surface is changed from 4000Å to 3000Å , so change in stopping potential will be……….

0.345 V

0.435 V

0.543 V

0.534 V

View Solution

Answer :

λ₁ = 4000Å = 4 × 10^–7 m

λ₂ = 3600Å = 3.6 × 10^–7 m

we know KE = (hc / λ) – ɸ

and KE = eV hence

eV = (hc / λ) – ɸ

hence V = (1/e) [(hc / λ) – ɸ]

change in stopping potential = V₂ – V₁

& V₂ – V₁ = (hc / e) [(1 / λ₂) – ɸ – (1 / λ₁) + ɸ]

∴ V₂ – V₁ = (hc / e) [(1 / λ₂) – (1 / λ₁)]

∴ V₂ – V₁ = [{6.62 × 10^–34 × 3 × 10⁸} / {1.6 × 10^–19}] [(10⁷ / 3.6)

– (10⁷ / 4)]

= [{6.62 × 3} / {1.6}] × {(4 – 3.6) / (4 × 3.6)}

= 12.41 × {(0.4) / (14.4)}

V₂ – V₁ = 0.344 V

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