Question

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, 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

$k = \frac{hc}{\lambda} - \Phi = E$
and,
$2k = \frac{hc}{\lambda^2} - \Phi = 2E$
⇒$$\frac{hc}{\lambda} - E = \frac{hc}{\lambda^2} - 2E
$⇒$ $\frac{hc}{\lambda^2} = \frac{hc}{\lambda} + E$
$⇒$ $\lambda^2 = \frac{hc\lambda}{hc + \lambda E}$
So, the correct option is (B).