The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \text{ m/s}$. The value of ratio of charge (e) to mass (m) of the photoelectron is $1.8 \times 10^{11} \text{ C/kg}$. The value of stopping potential in volt is
Show Hint
Stopping potential is independent of the number of electrons and depends only on their maximum speed.
Step 1: Understanding the Concept:
In the photoelectric effect, the stopping potential is the negative voltage required to bring the fastest-moving photoelectrons to a halt.
At the stopping potential, the electrical work done exactly equals the maximum kinetic energy of the emitted electrons. Step 2: Key Formula or Approach:
Equating work and kinetic energy: $e V_s = K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$.
We need to find the stopping potential $V_s$.
Rearranging the formula to utilize the given $e/m$ ratio: $V_s = \frac{v_{\text{max}}^2}{2(e/m)}$. Step 3: Detailed Explanation:
Given maximum velocity $v_{\text{max}} = 9 \times 10^5 \text{ m/s}$.
Given charge-to-mass ratio $\frac{e}{m} = 1.8 \times 10^{11} \text{ C/kg}$.
Using the energy equivalence equation:
\[ e V_s = \frac{1}{2} m v_{\text{max}}^2 \]
Isolate $V_s$:
\[ V_s = \frac{m v_{\text{max}}^2}{2e} = \frac{1}{2} \left(\frac{m}{e}\right) v_{\text{max}}^2 \]
Since we are given $e/m$, we can write this as division by $e/m$:
\[ V_s = \frac{v_{\text{max}}^2}{2 (e/m)} \]
Substitute the given values into the equation:
\[ V_s = \frac{(9 \times 10^5)^2}{2 \times (1.8 \times 10^{11})} \]
Square the velocity term:
\[ V_s = \frac{81 \times 10^{10}}{3.6 \times 10^{11}} \]
Adjust the powers of 10 to make division easier:
\[ V_s = \frac{8.1 \times 10^{11}}{3.6 \times 10^{11}} \]
The $10^{11}$ terms cancel out:
\[ V_s = \frac{8.1}{3.6} = \frac{81}{36} \]
Divide both numerator and denominator by 9:
\[ V_s = \frac{9}{4} = 2.25 \text{ V} \]
Step 4: Final Answer:
The stopping potential is $2.25 \text{ V}$.