Question

Light of wavelength $\lambda$ is incident on metal surface. Work function of metal is very small compare to kinetic energy of photon. If light of wavelength ${\lambda }^{\prime }$ is incident on metal surface then linear momentum of electron becomes $1.5$ of the momentum is previous case. Choose the correct value of ${\lambda }^{\prime }$:

• A. $\left(\frac{9\lambda }{4}\right)$
• B. $\left(\frac{3\lambda }{4}\right)$
• C. $\left(\frac{3\lambda }{2}\right)$
• D. $\left(\frac{4\lambda }{9}\right)$

Answer 1

Answer: (D)

$\left(\frac{4\lambda }{9}\right)$

$\frac{P^2}{2m}=\frac{hc}{\lambda }$
$\frac{(1.5P{)}^2}{2m}=\frac{hc}{{\lambda }^{\prime }}$
$\frac{1}{2.25}=\frac{{\lambda }^{\prime }}{\lambda }$
${\lambda }^{\prime }=\left(\frac{\lambda }{2.25}\right)=\left(\frac{4\lambda }{9}\right)$