Question

An electromagnetic wave of wavelength '${\displaystyle \lambda }$' is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength '${\displaystyle {\lambda }_1}$' then ____________.

Options

• ${\displaystyle \lambda \propto {\lambda }_1}$

• ${\displaystyle \lambda \propto \frac{1}{{\lambda }_1^2}}$

• ${\displaystyle \lambda \propto {\lambda }_1^2}$

• ${\displaystyle \lambda \propto \frac{1}{{\lambda }_1}}$

Answer 1

An electromagnetic wave of wavelength '${\displaystyle \lambda }$' is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength '${\displaystyle {\lambda }_1}$' then ${\displaystyle \lambda \propto {\lambda }_1^2}$.

Explanation:

${\displaystyle \frac{1}{2}{\text{mV}}^2=\frac{\text{hc}}{\lambda }}$

${\displaystyle \therefore \frac{{\text{p}}^2}{2\text{m}}=\frac{\text{hc}}{\lambda }}$

${\displaystyle \therefore {\text{p}}^2\propto \frac{1}{\lambda }}$

${\displaystyle \text{p}=\frac{\text{h}}{{\lambda }_1}}$

${\displaystyle \therefore \frac{{\text{h}}^2}{{\lambda }_1^2}\propto \frac{1}{\lambda }}$

${\displaystyle \therefore \lambda \propto {\lambda }_1^2}$