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: The energy of the \(H_\alpha\) photon is the difference between the \(n=3\) and \(n=2\) levels of hydrogen:
\[E = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right)\ \text{eV}\]

Step 2: Evaluate the bracket:
\[E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6\left(\frac{9-4}{36}\right) = 13.6\times\frac{5}{36}\ \text{eV}\]

Step 3: This gives
\[E = \frac{68}{36} \approx 1.89\ \text{eV} \approx 1.9\ \text{eV}\]

Step 4: Photoemission requires the photon energy to be at least equal to the work function, \(E \ge \phi\). The maximum allowed work function is therefore equal to the photon energy.

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