Photoelectric work function of a metal is 1 eV. If a light of wavelength of 500 Ao falls on the metal, the speed of the photoelectron emitted approximately will be: ( h = 6.625x 10-34 J.s)

hv=w+KE=w+12mev2 Or hcλ=1+12mev2 Or 6.625×10−34×3×108500×10−10 =1.6×10−19+12×9.1×10−31×v2 Or 12×9.1×10−31×v2 =6.625×35×10−18−1.6×10−19=38.15×10−19 Or v2=38.15×2×10129.1 Or v2=8.385×1012 Or v=2.89×106m/s∴ vc=3×1063×108=1100

Photoelectric work function of a metal is 1 eV. If a light of wavelength of 500 Ao falls on the metal, the speed of the photoelectron emitted approximately will be: ( h = 6.625x 10-34 J.s)

hv=w+KE=w+12mev2 Or hcλ=1+12mev2 Or 6.625×10−34×3×108500×10−10 =1.6×10−19+12×9.1×10−31×v2 Or 12×9.1×10−31×v2 =6.625×35×10−18−1.6×10−19=38.15×10−19 Or v2=38.15×2×10129.1 Or v2=8.385×1012 Or v=2.89×106m/s∴ vc=3×1063×108=1100

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Photoelectric work function of a metal is 1 eV. If a light of wavelength of 500 $\overset{o}{A}$ falls on the metal, the speed of the photoelectron emitted approximately will be: ( h = 6.625
x 10^-34 J.s)

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$\frac{1}{1000} \times velocity of light$

Equal to velocity of light

$\frac{1}{100} \times velocity of light$

$\frac{1}{10} \times velocity of light$

answer is C.

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Detailed Solution

$h v = w + KE = w + \frac{1}{2} m_{e} v^{2}$

$\begin{array}{l} Or h \frac{c}{λ} = 1 + \frac{1}{2} m_{e} v^{2} \\ Or \frac{6.625 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 10}} \end{array}$

$\begin{array}{l} = 1.6 \times 10^{- 19} + \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} \\ Or \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} \\ = \frac{6.625 \times 3}{5} \times 10^{- 18} - 1.6 \times 10^{- 19} = 38.15 \times 10^{- 19} \\ Or v^{2} = \frac{38.15 \times 2 \times 10^{12}}{9.1} \\ Or v^{2} = 8.385 \times 10^{12} \\ Or v = 2.89 \times 10^{6} m / s \\ ∴ \frac{v}{c} = \frac{3 \times 10^{6}}{3 \times 10^{8}} = \frac{1}{100} \end{array}$