DUAL NATURE OF MATTER AND RADIATION

1 )

Particle A and B have electric charge + q and + 4 q. Both have mass m. If both are allowed to fall under the same p.d., ratio of velocities (V_A / V_B) = …………

2:1

1:2

1:4

4:1

View Solution

Answer :

We know, Kinetic energy = work done

hence (1/2)mV² = q ∙ V where V is velocity, V is potential difference

∴ V =√{(2qV) / m}

here mass m, V is same hence

(V_A / V_B) = √(q_A / q_B) = √(q / 4_q) = (1/2) = 1:2

2 )

Energy of photon having wavelength λ is 2 eV. This photon when incident on metal. Maximum velocity of emitted is V. If λ is decreased 25% and maximum velocity is made double, work function of metal is……….eV

1.5

1.8

2.2

View Solution

Answer :

We know, (1/2) mV² = (hc / λ) – ɸ (1)

λ is decreased to λ¹ and λ¹ = 0.75λ. also V' = 2V

hence in later case.

(1/2)m (2V)² = {hc / (0.75λ)} – ɸ (2)

i.e. 4 × (1/2)mV² = (4/3) (hc / λ) – ɸ

from (1).

4{(hc / λ) – ɸ} = (4/3) (hc / λ) – ɸ

∴ (hc / λ) {4 – (4/3)} = 4ɸ – ɸ

∴ (hc / λ) (8/3) = 3ɸ

∴ ɸ = (hc / λ) (8/9)

given: (hc / λ) = 2 eV hence ɸ = 2 × (8/9) = 1.8 eV

3 )

When electric bulb having 100 W efficiency emits photon having wavelength 540 nm every second, numbers of photons will be……..(h = 6 × 10^–34 J.s, c = 3 × 10⁸ ms^–1)

3 × 10^(18)

3 × 10^(20)

100

View Solution

Answer :

P = {(nhc) / λ}

P = 100 W

h = 6 × 10^–34 J.s

c = 3 × 10⁸ m/s, λ = 540 nm = 540 × 10^–9 m

∴ n = no of photons = (Pλ / hc)

= {(100 × 540 × 10^–9) / (6 × 10^–34 × 3 × 10⁸)} = 3 × 10²⁰

4 )

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

increase by (λ/2)

decrease by (λ/2)

increase by 2λ

decrease 2λ

View Solution

Answer :

given: V₂ = 2V₁

KE of electron = E₁ = (1/2) mV₁² (1)

New KE of electron E₂ = (1/2) mV₂²

E₂ = (1/2)m (2V₁)²

∴ E₂ = 4[(1/2) mV₁²]

∴ E₂ = 4E₁ ------ from (1)

also

λ = [λ / {√2mE}]

As h and m is constant,

λ ∝ {1 / √E}

∴ (λ₂ / λ₁) = √(E₁ / E₂) = √{E₁ / 4E₁} = (1/2)

∴ λ₂ = (λ₁ / 2)

hence change in wavelength = λ₁ – λ₂

= λ₁ – (λ₁ / 2) = (λ₁ / 2)

i.e. decreases by (λ/2)

5 )

de-Broglie wavelength of proton accelerated under 100 V electric potential difference is λ₀. If de-wave length α-particle accelerated by the same electric potential difference what will its bouglie wavelength……….

2√2λ0

{λ0 / (2√2)}

(λ0 / √2)

(λ0 / 2)

View Solution

Answer :

Potential difference = 100V = constant

we know, (1/2) mV² = eV

i.e. mV² = 2eV

m²V² = 2meV²

but mV = p = momentum

∴ p = √(2meV)

also p = (h/λ) hence

(b/λ) = √(2meV)

here h₁, V is constant hence

λ ∝ {1 / (√me)}

(λ_∝ / λ_p) = √{(m_p e_p) / (m_∝ e_∝)}

m_∝ = 4mp

e_∝ = 2ep

hence (λ_∝ / λ_p) = √({m_p / (4m_p)} × {e_p / (2 e_p)})

= √{(1/4) × (1/2)}

(λ_∝ / λ_p) = {1 / (2√2)}

∴ λ_∝ = (λ_p / 2√2) = {λ₀ / (2√2)}

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