Question

When a light beam of frequency ‘ν’ is incident on a metal, photo electrons are emitted. If these electrons describe a circle of radius ‘r’ in a magnetic field of strength ‘B’, then the work function of the metal is:
$
  {\text{A}}{\text{. h}}\upsilon {\text{ + }}\dfrac{{{\text{Ber}}}}{{2{{\text{m}}_{\text{e}}}}} \\
  {\text{B}}{\text{. h}}\upsilon {\text{ - }}\dfrac{{{{\left( {{\text{Ber}}} \right)}^2}}}{{2{{\text{m}}_{\text{e}}}}} \\
  {\text{C}}{\text{. h}}\upsilon {\text{ + }}\dfrac{{{{\left( {{\text{Ber}}} \right)}^2}}}{{2{{\text{m}}_{\text{e}}}}} \\
  {\text{D}}{\text{. h}}\upsilon {\text{ - }}\dfrac{{{\text{Ber}}}}{{2{{\text{m}}_{\text{e}}}}} \\
$

Answer 1

- Hint: To find the work function of the metal we first determine the velocity of electrons described as a circle, in a magnetic field. Using the obtained velocity we calculate the kinetic energy of these electrons. Using the concept of photo electric effect we determine its work function.

Formula used,
Centripetal force of a system, $\dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$
Kinetic energy, $\dfrac{1}{2}{\text{m}}{{\text{v}}^2}$
In photoelectric effect, ${\text{h}}\upsilon {\text{ = }}{{\text{W}}_{\text{o}}} + {\text{ K}}{\text{.E}}$

Complete step-by-step solution:
When an electron is placed in a uniform magnetic field, it experiences a magnetic force on it. The magnetic force of an electron with a uniform magnetic field ‘B’ moving with a velocity ‘v’ is given by
F = Bev
This magnetic force on the electron makes it orbit in a circular path, the force keeping the electron in motion in a circular path is nothing but the centripetal force acting on it.
The centripetal force of an electron with mass ‘${{\text{m}}_{\text{e}}}$’, moving with a velocity ‘v’ along a radius ‘r’ is given by${\text{F = }}\dfrac{{{{\text{m}}_{\text{e}}}{{\text{v}}^2}}}{{\text{r}}}$.
This magnetic force provides the necessary centripetal force to the electron to keep it in the circular path.
Therefore we get,
${\text{Bev = }}\dfrac{{{{\text{m}}_{\text{e}}}{{\text{v}}^2}}}{{\text{r}}}$
$ \Rightarrow {\text{v = }}\dfrac{{{\text{eBr}}}}{{{{\text{m}}_{\text{e}}}}}$ ---- (1)

Now the kinetic energy of an electron of mass ‘${{\text{m}}_{\text{e}}}$’ moving with a velocity ‘v’ is given by
${\text{K}}{\text{.E = }}\dfrac{1}{2}{{\text{m}}_{\text{e}}}{{\text{v}}^2}$.
Using equation (1), K.E becomes
$
   \Rightarrow {\text{K}}{\text{.E = }}\dfrac{1}{2}{{\text{m}}_{\text{e}}}{\left( {\dfrac{{{\text{eBr}}}}{{{{\text{m}}_{\text{e}}}}}} \right)^2} \\
   \Rightarrow {\text{K}}{\text{.E = }}\dfrac{{{{\left( {{\text{Ber}}} \right)}^2}}}{{2{{\text{m}}_{\text{e}}}}}{\text{ - - - - - - (2)}} \\
$

We know, photoelectric effect states that
${\text{h}}\upsilon {\text{ = }}{{\text{W}}_{\text{o}}} + {\text{ K}}{\text{.E}}$, where ‘h’ is the planck's constant, ‘υ’ is the frequency of incident radiation, ${{\text{W}}_{\text{o}}}$is the work function and K.E is the kinetic energy.
Hence the work function is given as
$
   \Rightarrow {{\text{W}}_{\text{o}}} = {\text{h}}\upsilon {\text{ - K}}{\text{.E}} \\
   \Rightarrow {{\text{W}}_{\text{o}}} = {\text{h}}\upsilon {\text{ - }}\dfrac{{{{\left( {{\text{Ber}}} \right)}^2}}}{{2{{\text{m}}_{\text{e}}}}} \\
$ (Using equation (2))
Option B is the correct answer.

Note – In order to answer this type of question the key is to know the formula of photoelectric effect. Identifying the magnetic force on an electron only acts as the necessary centripetal force to keep it in the circular path is the key.
Work function is defined as the minimum amount of energy required to remove one electron from the metal. It depends on the atomic properties of the metal.