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:
We are given photons with an energy of $E = 10\text{ eV}$ incident on a photosensitive metal surface. The metal has a threshold frequency of $\nu_0 = 2 \times 10^{15}\text{ Hz}$. We need to calculate the maximum kinetic energy ($K.E._{\text{max}}$) of the emitted photoelectrons in electron-volts (eV).

Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation: $$K.E._{\text{max}} = E - \phi_0$$ where $E$ is the incident photon energy, and $\phi_0$ is the work function of the photosensitive metal surface. The work function can be calculated from the threshold frequency using: $$\phi_0 = h\nu_0$$ Since the final energy needs to be in eV, we must convert the work function from Joules to electron-volts using the conversion factor $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$.

Step 3: Detailed Explanation:
First, let's calculate the work function $\phi_0$ in Joules: $$\phi_0 = h\nu_0 = (6.63 \times 10^{-34}\text{ Js}) \times (2 \times 10^{15}\text{ Hz})$$ $$\phi_0 = 13.26 \times 10^{-19}\text{ J}$$ Now, convert this work function value into electron-volts (eV): $$\phi_0 = \frac{13.26 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} \approx 8.29\text{ eV}$$ Substitute the incident photon energy ($E = 10\text{ eV}$) and the calculated work function ($\phi_0 = 8.29\text{ eV}$) back into the photoelectric equation: $$K.E._{\text{max}} = 10\text{ eV} - 8.29\text{ eV} = 1.71\text{ eV}$$ This matches option (D).

Step 4: Final Answer:
The maximum kinetic energy of the emitted photoelectrons is $1.71\text{ eV}$, which corresponds to option (D).