Physics - Modern Physics Question with Solution | TestHub

Using Einstein's photoelectric equation:

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi$

When the wavelength is changed to $\frac{3\lambda }{4}$, the new maximum kinetic energy is:

$\frac{1}{2}m{v}^{\prime 2}=\frac{hc}{\frac{3\lambda }{4}}-\varphi =\frac{4hc}{3\lambda }-\varphi$

Substituting $\frac{hc}{\lambda }=\frac{1}{2}m{v}^2+\varphi$ into the above equation:

$\frac{1}{2}m{v}^{\prime 2}=\frac{4}{3}\left(\frac{1}{2}m{v}^2+\varphi \right)-\varphi$

$\frac{1}{2}m{v}^{\prime 2}=\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)+\frac{\varphi }{3}$

Since the work function $\varphi >0$, it follows that:

$\frac{1}{2}m{v}^{\prime 2}>\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)$

$v^{\prime 2}>\frac{4}{3}{v}^2$

$v^{\prime }>v{\left(\frac{4}{3}\right)}^{1/2}$

Answer: greater than $v{\left(\frac{4}{3}\right)}^{1/2}$