A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A…

A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A and B with monochromatic radiation of various frequencies and records the value of corrosponding stopping potentials $\displaystyle \left(\mathrm{V}_{\mathrm{s}}\right)$. The graph shows the variation of stopping potential $\displaystyle \left(\mathrm{V}_{\mathrm{s}}\right)$ with the frequency of incident radiation ( $\displaystyle \mathrm{v}$ ) for metals A and B.

Answer the following questions :

From the graph, the work functions of A and B are (h is Planck's constant and e value of charge on an electron)

$\displaystyle \mathrm{v}_{1}$ and $\displaystyle \mathrm{v}_{2}$

$\displaystyle \mathrm{V}_{1}$ and $\displaystyle \mathrm{V}_{2}$

$\displaystyle \mathrm{h} \mathrm{v}_{1}$ and $\displaystyle \mathrm{h} \mathrm{v}_{2}$

$\displaystyle \frac{\mathrm{h} \mathrm{v}_{1}}{\mathrm{e}}$ and $\displaystyle \frac{\mathrm{h} \mathrm{v}_{2}}{\mathrm{e}}$

For radiation of frequency $\displaystyle \mathrm{v}>\mathrm{v}_{2}$ incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

greater for metal A because it has a smaller work function.

greater for metal B because it has a larger work function.

greater for metal B because it has higher threshold frequency.

the same for both metal A and metal B because it is independent of work functions of metals.

If the intensity of the incident radiation for both metals A and B, is doubled keeping its frequency constant, then

the slope of the parallel lines will increase.

the slope of the parallel lines will decrease.

the threshold frequencies for both A and B will decrease.

the slope of the parallel lines will not change but more electrons will be emitted per second.

The threshold frequency for a metal surface is $\displaystyle \mathrm{v}_{0}$. If the radiation of frequency $\displaystyle 3 \mathrm{v}_{0}$ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is $\displaystyle \mathrm{E}_{1}$. If the frequency were increased to $\displaystyle 6 \mathrm{v}_{0}$, the maximum KE of the photoelectrons becomes $\displaystyle \mathrm{E}_{2}$. Then $\displaystyle \left(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right)$ equals

$\displaystyle 1$/$\displaystyle 3$

$\displaystyle 1$/$\displaystyle 2$

$\displaystyle 2$/$\displaystyle 5$

$\displaystyle 3$/$\displaystyle 4$

OR Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant ' $\displaystyle \mathrm{h}$ ' is given by

me

$\displaystyle \frac{1}{\mathrm{me}}$

$\displaystyle \frac{\mathrm{m}}{\mathrm{e}}$

$\displaystyle \frac{\mathrm{e}}{\mathrm{m}}$