Question

The radiation corresponding to $3 \to 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3 \times {10^{ - 4}}T$. If the radius of the largest circular path followed by these electrons is $10.0mm$, the work function of the metal is close to
(A) $0.8{\text{eV}}$
(B) ${\text{1}}{\text{.6eV}}$
(C) $1.8{\text{eV}}$
(D) $1.1{\text{eV}}$

Answer 1

Hint: To solve this question, we need to use the formula of the radius of the circular path followed by a charged particle when it enters a magnetic field. From there we can calculate the maximum kinetic energy of the emitted photoelectrons. Finally, using Einstein's photoelectric equation we can get the required value of the work function.
Formula used: The formulae used for solving this question are given by
$E = 13.6\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{eV}}$, here $E$ is the energy of the radiation emitted due to the transition ${n_2} \to {n_1}$ of an electron.
\[{K_{\max }} = h\nu - {\varphi _0}\], here ${K_{\max }}$ is the maximum kinetic energy of the photoelectrons, which are emitted by the incident light of frequency$\nu $on a metal surface having the work function${\varphi _0}$.
$r = \dfrac{{mv}}{{qB}}$, here $r$ is the radius of the circular path followed by a charged particle of mass $m$ and of charge $q$ when it enters in a magnetic field of $B$ with a velocity of $v$.

Complete step-by-step solution:
We know that the radius of the circular path followed by a charged particle when it enters in a magnetic field is given by
$r = \dfrac{{mv}}{{qB}}$
For an electron, we have $q = e$. Therefore we get
$r = \dfrac{{mv}}{{eB}}$
$ \Rightarrow mv = eBr$
Taking square both sides, we have
${m^2}{v^2} = {e^2}{B^2}{r^2}$
Dividing both sides by $2m$
$\dfrac{{{m^2}{v^2}}}{{2m}} = \dfrac{{{e^2}{B^2}{r^2}}}{{2m}}$
$\dfrac{1}{2}m{v^2} = \dfrac{{{e^2}{B^2}{r^2}}}{{2m}}$
We know that the kinetic energy is $K = \dfrac{1}{2}m{v^2}$. So we have
$K = \dfrac{{{e^2}{B^2}{r^2}}}{{2m}}$
Now, as we can see from the above relation, when the radius will be maximum, then the kinetic energy will also be maximum, that is,
${K_{\max }} = \dfrac{{{e^2}{B^2}{r_{\max }}^2}}{{2m}}$
According to the question, $B = 3 \times {10^{ - 4}}T$and ${r_{\max }} = 10mm = 0.01m$. Also we know that the mass of an electron is $m = 9.1 \times {10^{ - 31}}kg$ and the charge is $e = 1.6 \times {10^{ - 19}}C$. Substituting these above we get
${K_{\max }} = \dfrac{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}{{\left( {0.01} \right)}^2}}}{{2\left( {9.1 \times {{10}^{ - 31}}} \right)}}$
On solving we get
\[{K_{\max }} = 1.27 \times {10^{ - 19}}{\text{J}}\]
$ \Rightarrow {K_{\max }} = 0.8eV$ (1)
Now, we know that the energy of the radiation emitted due to the transition of electron is given by
$E = 13.6\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{eV}}$
According to the question, we have ${n_1} = 2$ and ${n_2} = 3$. Substituting these above we get
$E = 13.6\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right){\text{eV}}$
On solving we get
$E = 1.89{\text{eV}}$ (2)
From Planck’s law, we have
$E = h\nu $ (3)
Equating (2) and (3) we get
$h\nu = 1.89{\text{eV}}$ (4)
Now, from the Einstein’s photoelectric equation, we know that
\[{K_{\max }} = h\nu - {\varphi _0}\]
\[ \Rightarrow {\varphi _0} = h\nu - {K_{\max }}\]
Substituting (1) and (4)
\[{\varphi _0} = 1.89{\text{eV}} - 0.8{\text{eV}}\]
$ \Rightarrow {\varphi _0} = 1.09{\text{eV}} \approx {\text{1}}{\text{.1eV}}$
Thus, the work function of the metal is equal to $1.1{\text{eV}}$.

Hence, the correct answer is option D.

Note: We shouldn’t get confused by the phrase “largest circular path” given in the question. The largest circular path means the circular path which has the largest radius. The largest circular path will be followed by the photoelectrons which have the maximum kinetic energy.