Dual Nature of Matter

JEE Main level, single correct, medium

92 of 200

Question 92 of 200

Light source having wavelength 331331 nm is used to generate photo-electrons whose stopping potential is 0.20.2 V. The work function of the used metal in the experiment is α×10−19\alpha \times 10^{-19} J. The value of α\alpha is _____.
(h=6.62×10−34h = 6.62 \times 10^{-34} J s, e=1.6×10−19e = 1.6 \times 10^{-19} C and c=3×108c = 3 \times 10^8 m/s)

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Answer: (C) 5.685.68

Using Einstein's photoelectric equation:

hcλ=Φ+eV0\dfrac{hc}{\lambda} = \Phi + eV_0

The energy of the incident photon is:

E=hcλ=6.62×10−34×3×108331×10−9E = \dfrac{hc}{\lambda} = \dfrac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}

E=19.86×10−26331×10−9=6×10−19E = \dfrac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 6 \times 10^{-19} J

The maximum kinetic energy of the photoelectrons is:

Kmax=eV0=1.6×10−19×0.2=0.32×10−19K_{\text{max}} = eV_0 = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19} J

The work function Φ\Phi is:

Φ=E−Kmax\Phi = E - K_{\text{max}}

Φ=6×10−19−0.32×10−19=5.68×10−19\Phi = 6 \times 10^{-19} - 0.32 \times 10^{-19} = 5.68 \times 10^{-19} J

Comparing with α×10−19\alpha \times 10^{-19} J, we get α=5.68\alpha = 5.68.

Answer: 5.685.68

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