Question

Two streams of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of the ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $x:3$. The value of x is ________.

Answer 1

Step 1: Given

Energy of first photon: $E_1=\varphi$

Energy of second photon: $E_2=\varphi$

Ratio of maximum velocities of photons: $\frac{v_1}{v_2}=\frac{x}{3}$

where $v_1,v_{2}arevelocitiesoftwophoton$

Step 2: Formula Used

Photoelectric law: $K{E}_{max}=E-\varphi$

where $\varphi isworkfunction$

$KE=\frac{1}{2}m{v}^2$

Step 3: Calculate the ratio of velocities of the 2 photons

Calculate the velocity of the first electron using the photoelectric law

$$\begin{gathered} K{E}_{ma{x}_1}=E_1-\varphi \\ \Rightarrow \frac{1}{2}m{v_1}^2=2\varphi -\varphi \\ \Rightarrow {v_1}^2=\frac{2\left(2\varphi -\varphi \right)}{m} \\ \Rightarrow {v_1}^2=\frac{2\varphi }{m} \\ \Rightarrow v_1=\sqrt{\frac{2\varphi }{m}} \end{gathered}$$

Calculate the velocity of the second electron using the photoelectric law

$$\begin{gathered} K{E}_{ma{x}_2}=E_2-\varphi \\ \Rightarrow \frac{1}{2}m{v_2}^2=10\varphi -\varphi \\ \Rightarrow {v_2}^2=\frac{2\left(10\varphi -\varphi \right)}{m} \\ \Rightarrow {v_2}^2=\frac{18\varphi }{m} \\ \Rightarrow v_2=\sqrt{\frac{18\varphi }{m}} \end{gathered}$$

Calculate the ratio of the velocities by dividing the velocity of first photon by other

$$\begin{gathered} \frac{v_1}{v_2}=\frac{\sqrt{\frac{2\varphi }{m}}}{\sqrt{\frac{18\varphi }{m}}} \\ =\frac{\sqrt{1}}{\sqrt{9}} \\ =\frac{1}{3} \end{gathered}$$

Step 4: Find the value of x

Compare the values of the ratio given to the ratio obtained to find x

$$\begin{gathered} \frac{x}{3}=\frac{1}{3} \\ \Rightarrow x=1 \end{gathered}$$

Hence, the value of x is 1.