Question

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $\phi = 2.5$ eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. ($h=6.63 \times 10^{-34}$ Js, $c=3 \times 10^8$ $ms^{-1}$)

• A. 1.3 V
• B. 1.9 V
• C. 0.6 V
• D. 1.1 V

Hint:

Using $hc = 1240$ or $1242$ eV$\cdot$nm is much faster than multiplying individual constants in SI units and then converting.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Einstein's photoelectric equation relates incident energy, work function, and stopping potential. Changing the wavelength changes the incident energy, thereby changing the stopping potential.
Step 2: Key Formula or Approach:
\[ eV_s = \frac{hc}{\lambda} - \phi \]
where \(hc \approx 1240 \text{ eV}\cdot\text{nm}\).
Step 3: Detailed Explanation:
1. For \(\lambda_1 = 280 \text{ nm}\):
\[ E_1 = \frac{1240}{280} \text{ eV} \approx 4.43 \text{ eV} \]
\[ eV_{s1} = 4.43 - 2.5 = 1.93 \text{ eV} \implies V_{s1} = 1.93 \text{ V} \]
2. For \(\lambda_2 = 400 \text{ nm}\):
\[ E_2 = \frac{1240}{400} \text{ eV} = 3.10 \text{ eV} \]
\[ eV_{s2} = 3.10 - 2.5 = 0.60 \text{ eV} \implies V_{s2} = 0.60 \text{ V} \]
3. Change in stopping potential:
\[ \Delta V_s = 1.93 - 0.60 = 1.33 \text{ V} \approx 1.3 \text{ V} \]
Step 4: Final Answer:
The change in stopping potential is 1.3 V.