JEE Main 2019 — Structure Of Atom Question with Solution

From photoelectric effect,

E = $$\varphi$$ + KE

$$\frac{hc}{\lambda }=\varphi +\frac{p^2}{2m}$$ ................ (1)

Now when momentum = 1.5p then let wavelength = $${\lambda }_1$$

$$\therefore$$ $$\frac{hc}{{\lambda }_1}=\varphi +\frac{{\left(1.5p\right)}^2}{2m}$$ ........... (2)

Given,

kinetic energy(KE) of ejected photoelectron to be very high in comparison to work function($$\varphi$$).

$$\therefore$$ We can neglect work function($$\varphi$$).

$$\therefore$$ Equation (1) and (2) becomes,

$$\frac{hc}{\lambda }=\frac{p^2}{2m}$$ ................ (1)

$$\frac{hc}{{\lambda }_1}=\frac{{\left(1.5p\right)}^2}{2m}$$ ........... (2)

Dividing (1) by (2) we get,

$$\frac{{\lambda }_1}{\lambda }=\frac{p^2}{{\left(1.5p\right)}^2}$$

$$\Rightarrow$$ $$\frac{{\lambda }_1}{\lambda }=\frac{4}{9}$$

$$\Rightarrow$$ $${\lambda }_1=\frac{4}{9}\lambda$$