Question

Photons of energy $10\text{ eV}$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15}\text{ Hz}$. The kinetic energy in eV of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34}\text{ Js}$]

• A. $8.29\text{ eV}$
• B. $6.5\text{ eV}$
• C. $4.2\text{ eV}$
• D. $1.71\text{ eV}$

Hint:

To save time during competitive exams, memorize the value of Planck's constant directly in units of $\text{eV}\cdot\text{s}$: $h \approx 4.14 \times 10^{-15}\text{ eV}\cdot\text{s}$. Multiplying this directly by the threshold frequency gives $\phi_0 = (4.14 \times 10^{-15}) \times (2 \times 10^{15}) = 8.28\text{ eV}$ in one single line without dealing with large Joules conversions!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Photons of energy 10 eV strike a metal with threshold frequency 2×10¹⁵ Hz; find maximum K.E. of photoelectrons in eV.

Step 2: Key Formula or Approach:
K.E._max = E – φ₀, where work function φ₀ = hν₀. Convert φ₀ from joules to eV (÷ 1.6×10⁻¹⁹).

Step 3: Detailed Explanation:
φ₀ = (6.63×10⁻³⁴)(2×10¹⁵) = 13.26×10⁻¹⁹ J = 8.29 eV. K.E._max = 10 – 8.29 = 1.71 eV.

Step 4: Final Answer:
Maximum kinetic energy is 1.71 eV, matching option (D).