Fig- shows graphs between cut-off voltage V 0 and 1- for three metals 1-2 and 3 - where - is the wavelength of the incident radiation in nm- If W 1- W 2 and W 3 are the work functions of metals 1-2 and 3 respectively- thenA- W 1- W 2- W 3-1- 2- 4B- W 1- W 2- W 3-4- 2- 1C- The graphs for metals 1- 2 and 3 are parallel to each other and the slope of each graph is hc-e- where h - Planck-s constant- c - speed of light and e - charge of an electronD- Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from metal 3 -

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Photoelectric Effect

Fig. shows gr...

Question

Fig. shows graphs between cut-off voltage $V_{0}$ and $\frac{1}{λ}$ for three metals 1, 2 and 3, where $λ$ is the wavelength of the incident radiation in nm.
If $W_{1}, W_{2}$ and $W_{3}$ are the work functions of metals 1, 2 and 3 respectively, then

$W_{1} : W_{2} : W_{3} = 1 : 2 : 4$

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$W_{1} : W_{2} : W_{3} = 4 : 2 : 1$

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The graphs for metals 1, 2 and 3 are parallel to each other and the slope of each graph is hc/e, where h = Planck’s constant, c = speed of light and e = charge of an electron

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Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from metal 3.

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Solution

The correct options are
A
$W_{1} : W_{2} : W_{3} = 1 : 2 : 4$

C
The graphs for metals 1, 2 and 3 are parallel to each other and the slope of each graph is hc/e, where h = Planck’s constant, c = speed of light and e = charge of an electron

Work function $W = h v_{0} = \frac{h c}{λ_{0}}$ , where $λ_{0}$ is the threshold wavelength. Hence

$W_{1} : W_{2} : W_{3} = {\frac{h c}{(λ_{0})}}_{1} : {\frac{h c}{(λ_{0})}}_{2} : {\frac{h c}{(λ_{0})}}_{3} = 0.001 : 0.002 : 0.004 = 1 : 2 : 4$

Hence choice (a) is correct. In photoelectric emission, the relation between $V_{0}$ and $λ$ is given by

$e V_{0} = h ν - W = \frac{h c}{λ} - W$

or $V_{0} = \frac{h c}{e} (\frac{1}{λ}) - \frac{W}{e}$

Hence the slope of the graph between $V_{0}$ and $\frac{1}{λ}$ is $\frac{h c}{e}$ which is the same for all metals. Therefore, choice (c) is correct. The threshold wavelength for the three metals are

$\frac{1}{(λ_{0})_{1}} = 0.001 n m^{- 1}$ , therefore $(λ_{0})_{1} = 1000 n m = 10000^{\circ}$

$\frac{1}{(λ_{0})_{2}} = 0.002 n m^{- 1}$ , therefore $(λ_{0})_{2} = 500 n m = 5000^{\circ}$

$\frac{1}{(λ_{0})_{3}} = 0.004 n m^{- 1}$ , therefore $(λ_{0}) = 250 n m = 2500^{\circ}$

For photoelectric emission, the wavelength of the incident radiation must be less than the threshold wavelength. Since the wavelength of ultraviolet light is about $1200^{\circ}$ , it will eject photoelectrons from all the three metals. Hence the correct choices are (a) and (c).

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Fig. shows graphs between cut-off voltage V0 and 1λ for three metals 1, 2 and 3, where λ is the wavelength of the incident radiation in nm. If W1,W2 and W3 are the work functions of metals 1, 2 and 3 respectively, then

Fig. shows graphs between cut-off voltage $V_{0}$ and $\frac{1}{λ}$ for three metals 1, 2 and 3, where $λ$ is the wavelength of the incident radiation in nm.
If $W_{1}, W_{2}$ and $W_{3}$ are the work functions of metals 1, 2 and 3 respectively, then