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Q.

The K.E of an electron is E, when the wavelength of incident radiation is $λ$ . To increase the K.E. of electron to 2E, the incident wavelength must be:

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a

$\frac{hcλ}{Eλ + hc}$

b

$\frac{λ}{2}$

c

$2 λ$

d

$\frac{hcλ}{Eλ - hc}$

answer is B.

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Detailed Solution

$\frac{λ_{1}}{λ_{2}} = \frac{{KE}_{2}}{{KE}_{1}} λ_{2} = \frac{λ_{1} \times {KE}_{1}}{{KE}_{2}} = \frac{λE}{2 E} = \frac{λ}{2}$

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