Question

Work function \(\phi\) of a metal is \(9.53 \times 10^{-19}\) J, and the frequency of incident light is \(3.5 \times 10^{16}\) Hz. Find the kinetic energy of photoelectron. (Given Planck's constant \(h = 6.62 \times 10^{-34}\) Js) ?

• A. \(2.8 \times 10^{-17}\) J
• B. \(5.2 \times 10^{-17}\) J
• C. \(2.2 \times 10^{-17}\) J
• D. \(9.1 \times 10^{-17}\) J

Hint:

For photoelectric effect problems: \[ K.E.=h\nu-\phi \] Always calculate photon energy first and then subtract the work function.

Answer 1

The Correct Option is C

Concept: The photoelectric effect explains the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. According to Einstein’s photoelectric equation: :contentReference[oaicite:0]{index=0} where:
• \(K.E.\) = maximum kinetic energy of emitted photoelectron
• \(h\) = Planck’s constant
• \(\nu\) = frequency of incident radiation
• \(\phi\) = work function of metal

Step 1: Write all the given quantities clearly. The given values are: \[ \phi = 9.53 \times 10^{-19}\ \text{J} \] \[ \nu = 3.5 \times 10^{16}\ \text{Hz} \] \[ h = 6.62 \times 10^{-34}\ \text{Js} \] We are required to calculate the kinetic energy of the emitted photoelectron.

Step 2: Calculate the energy of the incident photon. Using: \[ E = h\nu \] Substituting the values: \[ E = (6.62 \times 10^{-34})(3.5 \times 10^{16}) \] Now multiply the numerical values: \[ 6.62 \times 3.5 = 23.17 \] and combine the powers of ten: \[ 10^{-34} \times 10^{16} = 10^{-18} \] Therefore: \[ E = 23.17 \times 10^{-18}\ \text{J} \] Rewriting in standard scientific notation: \[ E = 2.317 \times 10^{-17}\ \text{J} \]

Step 3: Apply Einstein’s photoelectric equation. Using: \[ K.E. = h\nu - \phi \] Substitute the calculated values: \[ K.E. = 2.317 \times 10^{-17} - 9.53 \times 10^{-19} \] Convert both quantities to same power: \[ 9.53 \times 10^{-19} = 0.0953 \times 10^{-17} \] Hence: \[ K.E. = 2.317 \times 10^{-17} - 0.0953 \times 10^{-17} \] \[ K.E. = 2.2217 \times 10^{-17}\ \text{J} \] Approximating: \[ K.E. \approx 2.2 \times 10^{-17}\ \text{J} \]

Step 4: Write the final answer. Therefore, the kinetic energy of the emitted photoelectron is: \[ \boxed{2.2 \times 10^{-17}\ \text{J}} \] Hence, the correct option is: \[ \boxed{(C)} \]