Question

Photoelectrons are emitted with maximum velocity $v$ when light of frequency $3f$ incidents on a photosensitive material of work function $2hf$. If the frequency of the incident light is $4.25 f$, the maximum velocity of the emitted photoelectrons is (h -- Planck's constant)

• A. $0.5 \text{ v}$
• B. $v$
• C. $1.5 \text{ v}$
• D. $2 \text{ v}$

Hint:

Photoelectric Principle:

• $K = h\nu - \phi$
• Kinetic energy is related to $v$ via $K = \frac12mv^2$
• For comparison, set up both kinetic energies and take their ratio
• Always use absolute frequency in $hf$ form, compare final vs initial

Answer 1

The Correct Option is C

According to Einstein’s photoelectric equation: \[ K = h\nu - \phi = \frac{1}{2}mv^2 \] For frequency $3f$: \[ \frac{1}{2}mv^2 = h(3f) - 2hf = hf \Rightarrow hf = \frac{1}{2}mv^2 \] Now, for frequency $4.25f$: \[ K' = h(4.25f) - 2hf = 2.25hf = \frac{1}{2}m(v')^2 \] Substitute $hf = \frac{1}{2}mv^2$: \[ \frac{1}{2}m(v')^2 = 2.25 \cdot \frac{1}{2}mv^2 \Rightarrow (v')^2 = 2.25v^2 \Rightarrow v' = \sqrt{2.25}v = 1.5v \] Thus, the new velocity is $1.5v$, matching option (3).

Answer 2

Step 4: Express velocity v in terms of kinetic energy
K.E._max = (1/2) m v² = hf

Step 5: Calculate K.E. for frequency 4.25f
Energy of photon at 4.25f: E = h × 4.25f = 4.25hf
Work function φ = 2hf (same as before)
Maximum kinetic energy at 4.25f:
K.E._max = 4.25hf − 2hf = 2.25hf

Step 6: Find the new maximum velocity v'
Let the new maximum velocity be v'. Then:
(1/2) m v'² = 2.25hf

Step 7: Relate v' to v
From Step 4, hf = (1/2) m v²
Divide the two equations:
(v'²) / (v²) = 2.25
=> v' / v = √2.25 = 1.5

Step 8: Conclusion
The maximum velocity of the photoelectrons when frequency is 4.25f is 1.5 times the velocity v.
Hence, the answer is 1.5 v.