Question

The light emitted in the transition \(n = 3\) to \(n = 2\) in hydrogen is called \(H_\alpha\) light. The maximum work function a metal can have so that \(H_\alpha\) light can emit photoelectrons from it is:

• A. \(3\ \text{eV}\)
• B. \(1.9\ \text{eV}\)
• C. \(5.1\ \text{eV}\)
• D. \(7.5\ \text{eV}\)

Hint:

Compute the \(n=3\to2\) energy \(13.6(1/4-1/9)\ \text{eV}\); the max work function equals this photon energy.

Answer 1

The Correct Option is B

Step 1: A metal emits photoelectrons only when the incident photon carries at least the binding (work-function) energy. So the ceiling on the work function equals the $H_\alpha$ photon energy.

Step 2: The $H_\alpha$ line is the Balmer transition from $n=3$ to $n=2$. Using $E_n = -\dfrac{13.6}{n^2}\ \text{eV}$, the emitted energy is $E_3 - E_2$ in magnitude:
\[|E| = 13.6\left(\frac{1}{4} - \frac{1}{9}\right)\ \text{eV}\]

Step 3: Numerically $E_2 = -3.40\ \text{eV}$ and $E_3 = -1.51\ \text{eV}$, so the photon energy is $-1.51 - (-3.40) = 1.89\ \text{eV} \approx 1.9\ \text{eV}$.

Step 4: Setting the threshold condition $\phi_{max} = E_{photon}$ gives the largest work function for which emission is still just possible.

Step 5: Therefore
\[\boxed{\phi_{max} = 1.9\ \text{eV}}\]