Question

When photons incident on a photosensitive material of work function 1.5 eV, the maximum velocity of the emitted photoelectrons is $8\times10^5$ ms$^{-1}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9\times10^{-31}$ kg and charge of the electron $= 1.6\times10^{-19}$ C)

• A. 1.8 V
• B. 1.5 V
• C. 2.1 V
• D. 2.4 V

Hint:

The stopping potential $V_s$ is directly related to the maximum kinetic energy of photoelectrons by $K_{max} = eV_s$. $K_{max}$ can be found either from the electron's velocity ($1/2 mv_{max}^2$) or from Einstein's photoelectric equation ($K_{max} = hf - \phi$). Choose the appropriate formula based on the given information.

Answer 1

The Correct Option is A

The stopping potential ($V_s$) is the potential difference required to stop the most energetic photoelectrons.
The relationship between the stopping potential and the maximum kinetic energy ($K_{max}$) of the photoelectrons is:
$K_{max} = e V_s$.
Where $e$ is the elementary charge.
The maximum kinetic energy can also be calculated from the maximum velocity ($v_{max}$) of the photoelectrons:
$K_{max} = \frac{1}{2} m v_{max}^2$.
Where $m$ is the mass of the electron.
We are given:
$v_{max} = 8 \times 10^5$ m/s.
$m = 9 \times 10^{-31}$ kg.
$e = 1.6 \times 10^{-19}$ C.
First, calculate the maximum kinetic energy in Joules:
$K_{max} = \frac{1}{2} (9 \times 10^{-31} \text{ kg}) (8 \times 10^5 \text{ m/s})^2$.
$K_{max} = \frac{1}{2} (9 \times 10^{-31}) (64 \times 10^{10}) = 9 \times 32 \times 10^{-21} = 288 \times 10^{-21}$ J.
Now, we can find the stopping potential $V_s$ from $K_{max} = e V_s$:
$V_s = \frac{K_{max}}{e} = \frac{288 \times 10^{-21} \text{ J}}{1.6 \times 10^{-19} \text{ C}}$.
$V_s = \frac{288}{1.6} \times 10^{-2} = 180 \times 10^{-2} = 1.8$ V.
The work function (1.5 eV) is extra information not needed to find the stopping potential from the given velocity.