Question

The work function of Caesium is 2.14 eV.

(a) Find the threshold frequency for Caesium.

(b) If the photocurrent is brought to zero by a stopping potential of 0.86 V, find the wavelength of the incident light.

Hint:

Use \(\nu_0 = \phi/h\) for the threshold frequency. For the wavelength, photon energy equals the work function plus the stopping energy, then \(\lambda = hc/E\).

Answer 1

Step 1: Understanding the Concept:
The work function \(\phi\) is the least energy needed to free an electron from the metal. Light can eject electrons only if its photon energy is at least \(\phi\).
The frequency at which photon energy just equals \(\phi\) is the threshold frequency \(\nu_0\).

Step 2: Key Formula or Approach:
(a) \(\phi = h\nu_0\), so \(\nu_0 = \phi / h\).
(b) Photon energy \(= \phi + eV_0\), where \(V_0\) is the stopping potential, and \(E = hc/\lambda\).

Step 3: Part (a): Threshold frequency:
Convert the work function to joules: \(\phi = 2.14 \times 1.6 \times 10^{-19} = 3.424 \times 10^{-19}\ \text{J}\).
\[ \nu_0 = \frac{\phi}{h} = \frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.16 \times 10^{14}\ \text{Hz} \]

Step 4: Part (b): Wavelength of the light:
The stopping potential 0.86 V means the largest kinetic energy is \(eV_0 = 0.86\ \text{eV}\).
Photon energy \(= 2.14 + 0.86 = 3.00\ \text{eV} = 3.00 \times 1.6 \times 10^{-19} = 4.8 \times 10^{-19}\ \text{J}\).
\[ \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4.8 \times 10^{-19}} \approx 4.14 \times 10^{-7}\ \text{m} \]

Final Answer:
(a) Threshold frequency \(\approx 5.2 \times 10^{14}\ \text{Hz}\). (b) Wavelength \(\approx 4.1 \times 10^{-7}\ \text{m}\) (about 414 nm). \[ \boxed{\nu_0 \approx 5.2 \times 10^{14}\ \text{Hz},\quad \lambda \approx 4.1 \times 10^{-7}\ \text{m}} \]