Question

When the electromagnetic radiation of frequency 4*10^15 Hz and 6*10^15 Hz fall on a same metal in different experiment, the ratio of the maximum kinetic energy of electron liberated 1:3. What is the threshold frequency for the metal?

Answer 1

Answer on Question #50773 – Physics – Nuclear Physics

1. When the electromagnetic radiation of frequency $4*10^15$ Hz and $6*10^15$ Hz fall on a same metal in different experiments, the ratio of the maximum kinetic energy of electron liberated 1:3. What is the threshold frequency for the metal?

After subtracting one can get that

$$h\nu_1 - h\nu_2 = E_1 - E_2, \quad h(\nu_1 - \nu_2) = E_1 - E_1 / 3, \quad E_1 = \frac{3h}{2}(\nu_1 - \nu_2).$$

The work function is $A = h\nu_1 - E_1 = h\nu_1 - \frac{3h}{2}(\nu_1 - \nu_2) = \frac{h}{2}(3\nu_2 - \nu_1)$.

For the threshold frequency, $h\nu_{th} = A$.

Thus, the threshold frequency for the metal is $\nu_{th} = \frac{A}{h}$, $\nu_{th} = \frac{3\nu_2 - \nu_1}{2}$.

Let check the dimension: $[v_{th}] = \text{GHz}$.

Let evaluate the quantity: $\nu_{th} = \frac{3 \cdot 6 \cdot 10^{15} - 4 \cdot 10^{15}}{2} = 7 \cdot 10^{15} \, (\text{GHz})$.

Answer: $7 \cdot 10^{15} \, \text{GHz}$.

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