Question

The ratio of the accelerating potentials required to accelerate (i) an \(α\)-particle and (ii) a proton to have the same de Broglie wavelength associated with them is
(mass of \(α\) - particle = \(6.4\times 10^{-27}\) kg, mass of proton = \(1.6\times 10^{-27}\) kg)

• A. \(1:8\)
• B. \(1:4\)
• C. \(8:1\)
• D. \(4:1\)

Hint:

Use Einstein photoelectric equation for the two wavelengths and eliminate the work function.

Answer 1

The Correct Option is A

Step 1: Einstein's equation:
\(\frac{hc}{\lambda} = \phi + eV_0\). Write \(E = \frac{hc}{\lambda}\).

Step 2: Two equations:
For \(\lambda\): \(E = \phi + eV\). For \(6\lambda\): \(\frac E6 = \phi + \frac{eV}{12}\).

Step 3: Eliminate:
Subtract: \(E - \frac E6 = eV - \frac{eV}{12}\), so \(\frac{5E}{6} = \frac{11eV}{12}\) and \(eV = \frac{10E}{11}\).
Then \(\phi = E - eV = E - \frac{10E}{11} = \frac E{11}\).

Step 4: Threshold wavelength:
\(\phi = \frac{hc}{\lambda_0} = \frac{E}{11} = \frac{hc}{11\lambda}\), so \(\lambda_0 = 11\lambda\).

Final Answer:
The threshold wavelength is \(11\lambda\), option (C). \[ \boxed{11\lambda} \]