Question

Question 1

a)velocity of an electron emitted by a metal whose threshold frequency is 2.24×10¹⁴ s−1s−1 when it is exposed to visible light of wavelength 5.14×10⁻⁷ mm .

Find the Velocity

b)What is the de Broglie wavelength of an electron traveling at 1.61×10⁵ m/sm/s?

Find de Broglie wavelength

Answer 1

Question.1 Ans :-

(a). From Photo-electric effect expression :

1/2 mv² = hv - hv₀ ............(1)

Where, m= Mass of electron = 9.1 x 10^-31 kg

v = velocity of electron emitted

h = Planck's constant = 6.626×10⁻³⁴ kg⋅m²⋅s⁻¹

v = Frequency of light = C (Velocity of light) / Wavelength of light (λ) = 3.0 x 10⁸ m.s^-1 /5.14×10⁻⁷ m = 5.837 x 10¹⁴ s^-1  

v₀ = Threshold frequency = 2.24×10¹⁴ s⁻¹

Put all these values in equation (1)

v² = 2 h(v - v₀) / m

v² = 2x6.626×10⁻³⁴ kg⋅m²⋅s⁻¹ (5.837 x 10¹⁴ s^-1 - 2.24×10¹⁴ s⁻¹ ) / (9.1 x 10^-31 kg )

v² = 4.767 x 10^-19 kg⋅m²⋅s⁻² / 9.1 x 10^-31 kg

v = (0.5238 x 10¹² m²⋅s⁻² )^1/2

v = 7.237 x 10⁵ m/s

Therefore, velocity of photo-electron emiited = 7.24 x 105 m/s

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(b). de Broglie wavelength = λ = h/mv

λ = 6.626×10⁻³⁴ kg⋅m²⋅s⁻¹ / (9.1 x 10^-31 kg x 1.61×10⁵ m/s)

λ = 4.52 x 10^-9 m

Therefore, de Broglie wavelength of an electron traveling = 4.52 x 10-9 m