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Q.

The photoelectric threshold wavelength of a certain metal is $2750 Å$ . Calculate the maximum velocity of the electrons ejected from the metal by light with a wavelength $1800 Å$ .

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a

$9.2 \times 10^{6} {ms}^{- 1}$

b

$5.2 \times 10^{5} {ms}^{- 1}$

c

$9.2 \times 10^{5} {ms}^{- 1}$

d

$5.2 \times 10^{6} {ms}^{- 1}$

answer is D.

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Detailed Solution

If the maximum speed of ejected electrons is $v_{\max}$ . then

$\frac{1}{2} {mv}_{\max}^{2} = K_{m a x} = \frac{h c}{λ} - \frac{h c}{λ_{t h}} = \frac{12375}{1800} - \frac{12375}{2750} = 6.875 - 4.5 \approx 2.4 eV$

$\Rightarrow v_{\max} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 2.4 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} \approx 9.2 \times 10^{5} {ms}^{- 1}$

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