SATHEE: Modern Physics 2 Question 9

Modern Physics 2 Question 9

9. Surface of certain metal is first illuminated with light of wavelength $λ_{1} = 350 n m$ and then by light of wavelength $λ_{2} = 540 n - m$ . It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2 . The work function of the metal (in eV) is close to (energy of photon $= \frac{1240}{λ (in n - m)} e V$ )

(a) 5.6

(b) 2.5

(d) 1.4

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Solution:

Let maximum speed of photo electrons in first case is $v_{1}$ and maximum speed of photo electrons in second case is $v_{2}$

Assumption I if we assume difference in maximum speed in two cases is 2 then $v_{1} = v$ and $v_{2} = 3 v$

According to Einstein’s photo electron equation Energy of incident photon $=$ work function $+ K E$

$i.e. \frac{h c}{λ} = φ_{0} + \frac{1}{2} m v^{2}$

Where $h c = 1240 e V, λ$ is wavelength of light incident, $φ_{0}$ is work function and $v$ is speed of photo electrons.

$\begin{aligned} When & λ_{1} & = 350 n m \\ ∴ & \frac{h c}{350} & = φ_{0} + \frac{1}{2} m v^{2} \\ or & \frac{h c}{350} - φ_{0} & = \frac{1}{2} m v^{2} \\ when & λ_{2} & = 540 n m \\ ∴ & \frac{h c}{540} & = φ_{0} + \frac{1}{2} m (3 v^{2}) \\ ∴ & \frac{h c}{540} - φ_{0} & = (\frac{1}{2} m v^{2}) \times 9 \end{aligned}$

Now, we divide Eq. (i) by Eq. (ii), we get

$\begin{aligned} \frac{\frac{h c}{350} - φ_{0}}{\frac{h c}{540} - φ_{0}} & = \frac{\frac{1}{2} m v^{2}}{(\frac{1}{2} m v^{2}) \times 9} = \frac{1}{9} \\ or 9 \frac{h c}{350} - φ_{0} & = \frac{h c}{540} - φ_{0} \\ or & 8 φ_{0} & = h c \frac{9}{350} - \frac{1}{540} \\ or & φ_{0} & = \frac{1}{8} \times 1240 \frac{9 \times 540 - 350}{350 \times 540} = 3.7 e V . \end{aligned}$

No option given is correct.

Alternate Method

Assumption II If we assume velocity of one is twice in factor with second, then.

Let $v_{1} = 2 v$ and $v_{2} = v$

We know that from Einstein’s photoelectric equation, energy of incident radiation $=$ work function $+ K E$

or $\frac{h c}{λ} = φ + \frac{1}{2} m v^{2}$

Let when $λ_{1} = 350 n m$ then $v_{1} = 2 v$

and when $λ_{1} = 540 n m$ then $v_{2} = v$

$∴$ Above Eq. becomes

$\begin{aligned} \frac{h c}{λ_{1}} & = φ + \frac{1}{2} m (2 v^{2}) \\ or & \frac{h c}{λ_{1}} - φ & = \frac{1}{2} m \times 4 v^{2} \\ and & \frac{h c}{λ_{2}} & = φ + \frac{1}{2} m v^{2} \\ or & \frac{h c}{λ_{2}} - φ & = \frac{1}{2} m v^{2} \end{aligned}$

Now, we divide Eq. (i) by (ii) Eq.

$\frac{\frac{h c}{λ_{1}} - φ}{\frac{h c}{λ_{2}} - φ} = \frac{\frac{1}{2} m \times 4 v^{2}}{\frac{1}{2} m v^{2}} = 4$

$\frac{h c}{λ_{1}} - φ = \frac{4 h c}{λ_{2}} - 4 φ$

$\begin{aligned} or φ = \frac{1}{3} h c \frac{4}{λ_{2}} - \frac{1}{λ_{1}} & = \frac{1}{3} \times 1240 \frac{4 \times 350 - 540}{350 \times 540} \\ or φ & = 1.8 e V \end{aligned}$