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)

Question

The difference between threshold wavelengths for two metal surfaces $ A $ and $ B $ having work functions $ \phi_A = 9 \, \text{eV} $ and $ \phi_B = 4.5 \, \text{eV} $ is: \[ \text{(Given: } h c = 1242 \, \text{eV nm)} \]

Answer 1

Step 1: Use stopping potential in energy form:
Let $K_1 = eV$ and $K_2 = \frac{eV}{12}$. The energies of photons are in the ratio $\frac{1}{\lambda} : \frac{1}{6\lambda} = 6 : 1$.

Step 2: Solve:
Photon energies are $6x$ and $x$. Then $6x - \phi = K_1$ and $x - \phi = \frac{K_1}{12}$. Multiply the second by 12: $12x - 12\phi = K_1$. Equate: $6x - \phi = 12x - 12\phi$, so $11\phi = 6x$, $\phi = \frac{6x}{11}$.
The photon energy for $\lambda$ is $6x$, so $\phi = \frac{6x}{11}$ corresponds to a wavelength 11 times longer: $\lambda_0 = 11\lambda$ (C).

Final Answer:
$11\lambda$. \[ \boxed{11\lambda} \]

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)

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The Correct Option is A

Solution and Explanation

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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)