Question

If a photocell is illuminated with a radiation of $1240\mathring{\text{A}},$ then the stopping potential is found to be $8\text{eV}.$ The work function of the emitter and the threshold wavelength are-

[Take : $hc=12400\text{(eV)}\mathring{\text{A}}$]

• A. $1\text{eV};5200\mathring{\text{A}}$
• B. $2\text{eV};6200\mathring{\text{A}}$
• C. $3\text{eV};7200\mathring{\text{A}}$
• D. $4\text{eV};4200\mathring{\text{A}}$

Answer 1

The correct option is B $2\text{eV};6200\mathring{\text{A}}$

Given,

$\lambda =1240\mathring{\text{A}};V_0=8\text{V}$

From, Einstein's photoelectric equation,

${\varphi }_0=h\nu -e{V}_0$

The energy of the incident photon is,

$E=\frac{hc}{\lambda }=\frac{12400\mathring{A}}{1240\mathring{A}}\text{eV}=10\text{eV}$

$\therefore {\varphi }_0=10-8=2\text{eV}$

Let, ${\lambda }_0=$ Threshold wavelength

$\because {\varphi }_0=h{\nu }_0=\frac{hc}{{\lambda }_0}$

${\lambda }_0=\frac{hc}{{\varphi }_0}=\frac{12400\mathring{\text{A}}\text{eV}}{2\text{eV}}=6200\mathring{\text{A}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.