Question

A beam of light consisting of two wavelengths $420\text{nm}$ and $540\text{nm}$ falls on two metals $1$ and $2$ of work functions $4.0\text{eV}$ and $2.5\text{eV}$ respectively. If the intensities of the two wavelengths are equal, which of the graphs shown in figure represents the variation of photoelectric current $\left(i\right)$ with voltage $\left(V\right)$ is correct?
[Take $hc=1240\text{eV (nm)}$ ]

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given,

$({\varphi }_0{)}_1=4\text{eV};({\varphi }_0{)}_2=2.54\text{eV}{\lambda }_1=420\text{nm};{\lambda }_2=540\text{nm}$

The Work function of a metal is,

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

$\Rightarrow ({\lambda }_0{)}_1=\frac{hc}{({\varphi }_0{)}_1}=\frac{1240}{4}=310\text{nm}$

$\Rightarrow ({\lambda }_0{)}_2=\frac{hc}{({\varphi }_0{)}_1}=\frac{1240}{2.5}=496\text{nm}$

As we know, for photo electric emission,

$\text{Threshold wavelength}\left({\lambda }_0\right)<\text{incident wavelength}\left(\lambda \right)$

Hence, metal $1$ emits photoelectrons for both the given wavelengths ($420\text{nm}$ and $540\text{nm}$) but metal $2$ emits photoelectrons only for wavelengths $540\text{nm}$. Therefore, photoelectric current $i$ will be more for metal $1$ than the metal $2$.

Now for the stopping potentials of the metals, using Einstein's photoelectric equation,

$E={\varphi }_0+K{E}_{max}$

$E-{\varphi }_0=e{V}_0$

For metal $1:$

$e(V_0{)}_1=\frac{hc}{\lambda }-({\varphi }_0{)}_1$

$=\frac{1240\text{eV (nm)}}{540\text{(nm)}}-4.0=-1.7\text{eV}$

For metal $2:$

$e(V_0{)}_2=\frac{hc}{\lambda }-({\varphi }_0{)}_2$

$=\frac{1240\text{eV (nm)}}{540\text{(nm)}}-2.5=-0.2\text{eV}$

Hence, we get $(V_0{)}_1>(V_0{)}_2$

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