Question

In an experiment of Photoelectric effect graph for $K{E}_{max}$ of ejected photoelectron v/s $h\nu$ of incident light is plotted. The x-intercept of the graph is 3.1 eV. Then threshold wavelength $\left({\lambda }_0\right)$ of metal is:

• A. 310 nm
• B. 400 nm
• C. 4000 nm
• D. 3100 nm

Answer 1

The correct option is B 400 nm

We know, according to Einstein equation,
$K{E}_{max}=hv-\varphi$
Where, $\varphi$ is the work function, that is the minimum energy required to remove an electron from the metal surface.
At the x-intercept $h\nu =h{\nu }_0$ since $K{E}_{max}=0$
$\varphi =3.1eV$
$\Rightarrow 3.1=\frac{1240\left(eV\right)}{{\lambda }_0\left(nm\right)}$
${\lambda }_0\left(nm\right)=400nm$