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Question 1 a)velocity of an electron emitted by a metal whose threshold frequency is 2.24×1014 s−1s−1...

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

Solutions

Expert Solution

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

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