Embibe Experts Solutions for Exercise 2: EXERCISE 11.2

Consider the graph between stopping potential $\left({\mathrm{V}}_0\right)$ versus frequency of incident light $\left(\mathrm{f}\right)$ shown here for two different metals.

Which of the metals has higher value of work function?

Consider the graph between stopping potential $\left(V_0\right)$ versus frequency of incident light $\left(f\right)$ shown here for two different metals.

If work function of metal. $A$ is $k$ times the work function of metal $B$, what is the relation between ${\theta }_1$ and ${\theta }_2 ?$ 

In a photoelectric emission the stopping potential is plotted against the frequency $v$ of incident light. The resulting curve is a straight line which makes an angle $\theta$ with the $\nu$-axis. Then $\tan \theta$ will be equal to ( $\varphi =$ work function $)$

The stopping potential are ${\mathrm{V}}_1$ and ${\mathrm{V}}_2$ with incident lights of wavelengths ${\mathrm{\lambda }}_1$ and ${\mathrm{\lambda }}_2$ respectively. Then ${\mathrm{V}}_1-{\mathrm{V}}_2$ equal to

When a certain metallic surface is illuminated with light of wavelength $\lambda$, the stopping potential is $3V_0$. When the same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface is

A proton and an $\alpha$ -particle are accelerated through the same potential difference. The ratio of de-Broglie wavelength of the proton to that of the $\alpha$ particle will be
 

The work functions of silver and sodium are $4.6$ and $2.3 \mathrm{eV}$, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is

In a photoelectric experiment, a parallel beam of monochromatic light with the power of $200 \mathrm{W}$ is incident on a perfectly absorbing cathode of work function $6.25 \mathrm{eV}$. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $100 \%$. A potential difference of $500 \mathrm{V}$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F=n\times {10}^{–4} \mathrm{N}$ due to the impact of the electrons. The value of $n$ is? Mass of the electron $m_e=9\times {10}^{-31} \mathrm{kg}$ and $1.0 \mathrm{eV}=1.6\times {10}^{–19} \mathrm{J}$.