Question

Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio h/e is:
A. \[6.6 \times {10^{ - 15}}Js{C^{ - 1}}\]
B. \[4.12 \times {10^{ - 15}}Js{C^{ - 1}}\]
C. \[6.6 \times {10^{ - 34}}Js{C^{ - 1}}\]
D. \[4.12 \times {10^{ - 34}}Js{C^{ - 1}}\]

Answer 1

Hint: Use Maximum kinetic energy to get Stopping potential and form two equations.
Use $h\nu = h{\nu _0} + K{E_{max}}$ and $K{E_{max}} = e{V_0}$ to find $\dfrac{h}{e}$ . We need to find only approximate answers in case of competitive exams.

Complete Step by Step answer:
we know that photons of light have frequency-dependent energy given by the equation: $e = h\nu $
When a photocathode is hit by light of certain frequency or wavelength, the photons of light transfer their energy to the metal.
Also, we know that every metal ejects electrons if the electron is given a sufficient amount of energy. This minimum energy required is called the work function ${W_0}$ of the metal.
So if the energy of the photon is greater than the work function, electrons would be ejected. The extra energy $(h\nu - {W_0})$ is present in the electron as its kinetic energy.
This phenomenon is explained by the einstein photoelectric equation :
$h\nu = h{\nu _0} + K{E_{max}}$. Here $h{\nu _0}$ is the work function of the metal.

The electrons ejected with kinetic energy K can be decelerated by applying a negative potential at the Anode. If the potential of ${V_0}$ could stop even the highest kinetic energy electron from coming out of the plate, then ${V_0}$ is the stopping potential.
To remove all the kinetic energy from the electron, the Potential should supply an equal amount of opposite potential energy.
So ${V_0}$ and$K{E_{max}}$ can be related as :
$K{E_{max}} = e{V_0}$

Now let's answer the question:
From the photoelectric equation, we get
$h\nu = {W_0} + K{E_{max}}$
Since Kinetic energy and stopping potential have same value,
$h\nu = {W_0} + e{V_0}$
Since the question has given us wavelength instead of frequency, lets replace frequency with wavelength.
$h\nu = {W_0} + e{V_0}$
$h\dfrac{c}{\lambda } = {W_0} + e{V_0}$
Now let's divide this equation throughout by $e$ to get $\dfrac{h}{e}$
$\dfrac{h}{e}\dfrac{c}{\lambda } = \dfrac{{{W_0}}}{e} + {V_0}$
Now let's substitute the values given in the question into the equation. But for that we have to convert $0.18eV$ of Kinetic energy to Stopping potential.
one electron volt is the energy gained by an electron if accelerated using 1 volt potential. Here since the energy is 0.18 eV, the potential applied was 0.18V.

Now we can substitute to get two equations:
$\dfrac{h}{e}\dfrac{c}{{546nm}} = \dfrac{{{W_0}}}{e} + 0.18$
$\dfrac{h}{e}\dfrac{c}{{185nm}} = \dfrac{{{W_0}}}{e} + 4.6$
Now let's solve these equations to get the answer.
$\Rightarrow \dfrac{h}{e}\dfrac{c}{{185nm}} - 4.6 = \dfrac{h}{e}\dfrac{c}{{546nm}} - 0.18$
$\Rightarrow \dfrac{h}{e}\left( {\dfrac{c}{{185nm}} - \dfrac{c}{{546nm}}} \right) = 4.6 - 0.18$
\[\Rightarrow \dfrac{h}{e}\left( {\dfrac{1}{{185}} - \dfrac{1}{{546}}} \right) = \dfrac{{\left( {4.6 - 0.18} \right)}}{3}{10^{ - 17}}\]
\[\Rightarrow \dfrac{h}{e}\left( {5.4 \times {{10}^{ - 3}} - 1.83 \times {{10}^{ - 3}}} \right) = \dfrac{{\left( {4.42} \right)}}{3}{10^{ - 17}}\]
\[\Rightarrow \dfrac{h}{e} = 4.12 \times {10^{ - 15}}Js{C^{ - 1}}\].

Note: It would be useful if we remember the value of \[\dfrac{{hc}}{e} = 1.242 \times {10^{ - 7}}\] for quicker calculations in competitive exams. Also note that the word maximum kinetic energy is used because most of the kinetic energy gained by an electron from the photon is lost due to collisions within the metal.