Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation:

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

When the maximum kinetic energy is doubled, the new wavelength is ${\lambda }_1$:

$2E=\frac{hc}{{\lambda }_1}-\varphi$

Substituting the value of $E$ from the first equation into the second:

$2\left(\frac{hc}{\lambda }-\varphi \right)=\frac{hc}{{\lambda }_1}-\varphi$

$\frac{2hc}{\lambda }-2\varphi =\frac{hc}{{\lambda }_1}-\varphi$

$\frac{hc}{{\lambda }_1}=\frac{2hc}{\lambda }-\varphi$

Since the work function $\varphi >0$, we have:

$\frac{hc}{{\lambda }_1}<\frac{2hc}{\lambda }\Rightarrow {\lambda }_1>\frac{\lambda }{2}$

Also, since the initial kinetic energy $E>0$, we know that $\frac{hc}{\lambda }>\varphi$. We can rewrite the equation for ${\lambda }_1$ as:

$\frac{hc}{{\lambda }_1}=\frac{hc}{\lambda }+\left(\frac{hc}{\lambda }-\varphi \right)$

Since $\left(\frac{hc}{\lambda }-\varphi \right)>0$, it follows that:

$\frac{hc}{{\lambda }_1}>\frac{hc}{\lambda }\Rightarrow {\lambda }_1<\lambda$

Combining both inequalities, we get:

$\frac{\lambda }{2}<{\lambda }_1<\lambda$

Answer: ${\lambda }_1>\frac{\lambda }{2}<\lambda$