Question

Threshold wavelength of a photometal is 24800 Aº. If radiations of energies 1.5eV and 2eV are
incident on it. Ratio between maximum speeds of ejected electrons is

Answer 1

Answer:

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Answer 2

Answer:

The ratio of the maximum velocity of the ejected electrons is $\sqrt{\frac{2}{3}}$.

Explanation:

We will use the photoelectric equation to solve this question,

$\Phi=\Phi_0+K_m_a_x$        (1)

Ф=energy of incident radiation

Ф₀=work function

Kmax=maximum kinetic energy=1\2×m×V²max

First, we need to find the work function of the photo metal in eV i.e.

$\Phi_0=\frac{12400}{\lambda_0}$           (2)

The values given in the question are,

λ₀=24800 Aº

Ф₁=1.5eV

Ф₂=2eV

Now,

$\Phi_0=\frac{12400}{24800 }=0.5eV$            (3)

We can also write equation (1) as,

$\frac{1}{2}\times m\times V^2_m_a_x=\Phi-\Phi_0$        (4)

Now as per the question,

$\frac{\frac{1}{2}\times m\times V^2_1_m_a_x_}{\frac{1}{2}\times m\times V^2_2_m_a_x_}}=\frac{\Phi_1-\Phi_0}{\Phi_2-\Phi_0}$

$\frac{V^2_1_m_a_x}{V^2_2_m_a_x}=\frac{\Phi_1-\Phi_0}{\Phi_2-\Phi_0}$         (5)

By substituting the required values in equation (5) we get;

$\frac{V^2_1_m_a_x}{V^2_2_m_a_x}=\frac{1.5-0.5}{2-0.5}=\frac{1}{1.5}=\frac{10}{15}=\frac{2}{3}$

$\frac{V_1_m_a_x}{V_2_m_a_x}=\sqrt{\frac{2}{3}}$

Hence, the ratio of the maximum velocity of the ejected electrons is $\sqrt{\frac{2}{3}}$.