Solveeit Logo

Question

Question: The wavelength of incident radiation becomes one-fourth times, the factor by which maximum velocity ...

The wavelength of incident radiation becomes one-fourth times, the factor by which maximum velocity changes (assuming work function to be negligible)
A. Four times (approximately)
B. Two times (approximately)
C. Half (approximately)
D. One-fourth times (approximately)

Explanation

Solution

In order to solve this question, we will use the general photoelectric effect equation and firstly, we will make a relation between wavelength and velocity of photon and electron and then by changing the wavelength according to the question, we will figure out the change in velocity of the electron.

Formula used:
The general photoelectric effect equation is written as
hcλ=W+12mv2\dfrac{{hc}}{\lambda } = W + \dfrac{1}{2}m{v^2}
where, h,ch, c and λ\lambda are Planck’s constant, speed of light and wavelength for the photon. WW represent work function and m,vm, v represent mass and maximum velocity of the electron.

Complete step by step answer:
According to the question, we have given that work function should be assumed negligible which means W=0W = 0 now, initially if λ,v\lambda ,v be the wavelength of photon and maximum velocity of electron then, we have photoelectric effect equation as,
hcλ=12mv2(i)\dfrac{{hc}}{\lambda } = \dfrac{1}{2}m{v^2} \to (i)
Now, if new wavelength be λ\lambda ' then, according to the question, it became one-fourth of initial wavelength hence,
λ=λ4\lambda ' = \dfrac{\lambda }{4}
Let new velocity of the electron will be v’ then, equation will be written as
hcλ=12mv2(ii)\Rightarrow \dfrac{{hc}}{{\lambda '}} = \dfrac{1}{2}mv{'^2} \to (ii)

Now divide the equation (i) by (ii) we get,
hcλhcλ=12mv212mv2\Rightarrow \dfrac{{\dfrac{{hc}}{\lambda }}}{{\dfrac{{hc}}{{\lambda '}}}} = \dfrac{{\dfrac{1}{2}m{v^2}}}{{\dfrac{1}{2}mv{'^2}}}
On simplifying, we get
λλ=(vv)2\Rightarrow \dfrac{{\lambda '}}{\lambda } = {(\dfrac{v}{{v'}})^2}
Now putting the value of λ=λ4\lambda ' = \dfrac{\lambda }{4} we get,
14=(vv)2\Rightarrow \dfrac{1}{4} = {(\dfrac{v}{{v'}})^2}
on simplifying, we get
v=2v\therefore v' = 2v
So, velocity becomes two times that of initial velocity.

Hence, the correct option is B.

Note: It should be remembered that, here we have assumed that work function is negligible but in practical applications of photoelectric effect no metal has the zero work function that’s why we have said that velocity will become two times approximately not exactly, and this famous photoelectric effect was explained by Einstein and he also won nobel prize for this work.