When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively ? [Take $\frac{hc}{e}=1.24\times {10}^{-6}{\text{J m C}}^{-1}$].

The correct answer is 1.72 × 10^–7 m, 1.20 eV.

This problem uses Einstein's Photoelectric Equation. The key idea is that when light of frequency f (or wavelength λ) strikes a metal, the maximum kinetic energy of the emitted photoelectron equals the photon energy minus the work function (φ) of the metal:

$E_{\mathrm{k(max)}}=\frac{\mathrm{hc}}{\lambda }-\phi$

The stopping potential V is the minimum reverse potential needed to bring these fastest electrons to rest, so $E_{\mathrm{k(max)}}=eV$. This gives:

$eV=\frac{\mathrm{hc}}{\lambda }-\phi$

Note: The intensity of light only affects the number of photoelectrons emitted, NOT their maximum kinetic energy. So the stopping potential depends only on the wavelength, not the intensity. The "intensity half" detail in the problem is a deliberate red herring.

Setting Up the Two Equations:

Let λ₁ = λ be the wavelength of the first source. The second source has wavelength λ₂ = 4λ.

For the first source (stopping potential V₁ = 6.0 V):

$e\times 6.0=\frac{\mathrm{hc}}{\lambda }-\phi$     ... (1)

For the second source (stopping potential V₂ = 0.6 V, wavelength 4λ):

$e\times 0.6=\frac{\mathrm{hc}}{\mathrm{4\lambda }}-\phi$     ... (2)

Step 1 — Subtract equation (2) from equation (1) to eliminate φ:

$e(6.0-0.6)=\frac{\mathrm{hc}}{\lambda }-\frac{\mathrm{hc}}{\mathrm{4\lambda }}$

$5.4e=\frac{\mathrm{hc}}{\lambda }(1-\frac{1}{4})=\frac{\mathrm{hc}}{\lambda }\times \frac{3}{4}$

Step 2 — Solve for λ:

$\frac{\mathrm{hc}}{\mathrm{e\lambda }}=5.4\times \frac{4}{3}=7.2\text{V}$

$\lambda =\frac{\mathrm{hc}}{e}=\frac{1.24}{\times }7.2$

$\lambda =0.1722\times {10}^{-6}\text{m}$

$\lambda =1.72\times {10}^{-7}\text{m}$

This is in the ultraviolet range, which makes sense for photoelectric effect on metals.

Step 3 — Find the work function φ:

Substituting back into equation (1), where we found $\frac{\mathrm{hc}}{\mathrm{e\lambda }}=7.2\text{V}$, which means $\frac{\mathrm{hc}}{\lambda }=7.2\text{eV}$:

$\phi =\frac{\mathrm{hc}}{\lambda }-e{V}_1=7.2\text{eV}-6.0\text{eV}$

$\phi =1.2\text{eV}$

Verification with source 2: Photon energy from source 2 = 7.2/4 = 1.8 eV. Kinetic energy = 1.8 − 1.2 = 0.6 eV ⇒ Stopping potential = 0.6 V ✓

Final Answers:

• Wavelength of the first source: 1.72 × 10^–7 m

• Work function of the metal: 1.20 eV