Photoelectric effect is observed on a surface | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - hf_0 = h(f - f_0)$.Given frequenci

Vedclass · Accepted Answer

$5.75 	imes 10^8 	ext{ MHz}$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - hf_0 = h(f - f_0)$.Given frequencies are $f_1 = 5.5 	imes 10^8 	ext{ MHz}$ and $f_2 = 4.5 	imes 10^8 	ext{ MHz}$.The ratio of maximum kinetic energies is $\frac{K_1}{K_2} = \frac{1}{5}$.Substituting the values: $\frac{K_1}{K_2} = \frac{h(f_1 - f_0)}{h(f_2 - f_0)} = \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{5}$.$\frac{5.5 	imes 10^8 - f_0}{4.5 	imes 10^8 - f_0} = \frac{1}{5}$.$5(5.5 	imes 10^8 - f_0) = 4.5 	imes 10^8 - f_0$.$27.5 	imes 10^8 - 5f_0 = 4.5 	imes 10^8 - f_0$.$4f_0 = 23.0 	imes 10^8$.$f_0 = 5.75 	imes 10^8 	ext{ MHz}$.

Photoelectric effect is observed on a surface for frequencies $5.5 \times 10^8 \text{ MHz}$ and $4.5 \times 10^8 \text{ MHz}$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:5$,then the threshold frequency for the metal surface is:

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