The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct [2006]

Question

The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct? [2006]

Smart Achievers · Accepted Answer

(1, 3)

From $e V = \frac{h c}{λ} - ϕ$

$V = \frac{h c}{e λ} - \frac{ϕ}{e}$

At $V = 0$ , $ϕ_{1} : ϕ_{2} : ϕ_{3} = \frac{h c}{λ_{1}} : \frac{h c}{λ_{2}} : \frac{h c}{λ_{3}}$

$= 0.001 : 0.002 : 0.004 = 1 : 2 : 4$

By Einstein's photoelectric equation, $\frac{h c}{λ} - ϕ = e V$

$\Rightarrow V = \frac{h c}{e λ} - \frac{ϕ}{e}$ $\dots (i)$

Comparing equation (i) by $y = m x + c,$ we get the slope of the line, $m = \frac{h c}{e} = \tan θ$

From the graph of $V_{0}$ versus $\frac{1}{λ}$ , it is clear that,

$\frac{1}{λ_{0_{1}}} = 0.001 {(nm)}^{- 1}$ $∴$ $λ_{0_{1}} = \frac{1}{0.001} = 1000 nm$

$\frac{1}{λ_{0_{2}}} = 0.002 {(nm)}^{- 1}$ $∴$ $λ_{0_{2}} = 500 nm$ and $λ_{0_{3}} = 250 nm$

Violet colour light has wavelength 400 nm.

Therefore, this light will be unable to show photoelectric effect on plate 3 and can eject photoelectrons from metal-1 and metal-2.

Solution

Q.
The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct [2006]

1 $ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$

2 $ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$

3 $\tan θ \propto \frac{h c}{e}$

4 Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

Want Us to Email you About Special Offers & Updates?

Solution

Q. The graph between the stopping potential (V0) and (1λ) is shown in the figure. ϕ1, ϕ2 and ϕ3 are work functions, which of the following is/are correct [2006]

1 ϕ1:ϕ2:ϕ3=1:2:4

2 ϕ1:ϕ2:ϕ3=4:2:1