The stopping potential in an experiment on a photoelectric effect is 1.5V. What is the maximum kinetic energy of the photoelectrons emitted?

1

(a)

\(Max.K.E.={ K }_{ max }={ eV }_{ 0 }=e\times 1.5V=1.5eV\)

A photon and an electron have been same de-Broglie wavelength which one has higher total energy?

1

(a)

Total energy of an electron E_e=mc²
Total energy of a photon, \(E_{p}=\frac{h c}{\lambda}\)
de-Broglie wavelength of electron of mass m moving with velocity V
\(\lambda=\frac{h}{m v}\)
\(\Rightarrow \ m=\frac{h}{\lambda v}\)
\(\therefore \text { Energy of an electron, } E_{e}=m c^{2}=\frac{h c^{2}}{\lambda v}\)
\(\therefore \ \frac{E_{e}}{E_{p}}=\frac{\frac{h c^{2}}{\lambda v}}{\frac{h c}{\lambda}}=\frac{c}{v}\)
As c > > v, therefore the total energy of electron is more than the total energy of photon.

Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of incident radiations.

1

(a)

If wavelength of electromagnetic waves are doubled what will happen to energy of photon?

1

(a)

E = hv, 
= hc/\(\lambda \)
E \(\infty \) 1/\(\lambda \) energy of proton to half

Alkali metals are most suitable for photoelectric emission. Why?

1

(a)

Alkali metals have too low work functions. Even visible light can eject electrons from them.