Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation:

$E=W_0+K_{\text{max}}$

where $E$ is the energy of the incident radiation, $W_0$ is the work function, and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.

The maximum kinetic energy is related to the stopping potential $V$ by:

$K_{\text{max}}=eV$

The energy of the incident radiation is given by:

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

Equating the expressions for $E$:

$\frac{hc}{\lambda }=W_0+eV$

Solving for $\lambda$:

$\lambda =\frac{hc}{W_0+eV}$

Answer: $\frac{\text{hc}}{({\text{W}}_0+\text{eV})}$