Question

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

• A. $\frac{hc}{E\lambda - hc}$
• B. $\frac{hc\lambda}{E\lambda + hc}$
• C. $\frac{h\lambda}{E\lambda + hc}$
• D. $\frac{hc\lambda}{E\lambda - hc}$

Answer 1

Answer: (B)

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

Answer 2

E = $\frac{hc}{\lambda}$ – f …….(1)

2E = $\frac{hc}{\lambda'}$ – f ….…(2)

2E – E = hc$\left\lbrack \frac{1}{\lambda'} - \frac{1}{\lambda} \right\rbrack$ = E

E + $\frac{hc}{\lambda}$= $\frac{hc}{\lambda'}$

$\frac{E\lambda + hc}{\lambda}$ = $\frac{hc}{\lambda'}$

l¢ = $\left( \frac{hc\lambda}{E\lambda + hc} \right)$