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Question: The kinetic energy of the electron is E when the incident wavelength is l. To increase the K.E. of t...

The kinetic energy of the electron is E when the incident wavelength is l. To increase the K.E. of the electron to 2E, the incident wavelength must be-

A

2l

B

l/2

C

hcλEλ+hc\frac{hc\lambda}{E\lambda + hc}

D

None

Answer

hcλEλ+hc\frac{hc\lambda}{E\lambda + hc}

Explanation

Solution

E = hcλ\frac{hc}{\lambda} – f0 ;

2E = hcλ′−φ0\frac{hc}{\lambda'} - \varphi_{0} or 2hcλ−2φ0=hcλ′−φ0\frac{2hc}{\lambda} - 2\varphi_{0} = \frac{hc}{\lambda'} - \varphi_{0}

or hcλ′=2hcλ−φ0\frac{hc}{\lambda'} = \frac{2hc}{\lambda} - \varphi_{0}

or hcλ′=2hcλ+E−hcλ\frac{hc}{\lambda'} = \frac{2hc}{\lambda} + E - \frac{hc}{\lambda}

or hcλ′=E+hcλ\frac{hc}{\lambda'} = E + \frac{hc}{\lambda}

or hcλ′=Eλ+hcλ\frac{hc}{\lambda'} = \frac{E\lambda + hc}{\lambda}

or l¢ = hcλEλ+hc\frac{hc\lambda}{E\lambda + hc}