Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases.

OR

(i) Name the experiment which confirms the existence of the wave nature of electrons. Derive the expression for the de-Broglie wavelength of an electron moving under a potential difference of V volts. (ii) An electron and a proton have the same Kinetic Energy. Which of these particles has a shorter de-Broglie wavelength?

1) Photoelectric current depends only on the intensity of photons and not on their frequency. So photoelectric current doesn't change when the frequency is doubled.

2) The kinetic energy of the emitted photons are given by KE = hν-hν₀

where ν = frequency of incident radiation

ν₀ = Threshold frequency of incident radiation.

As the frequency of incident radiation doubles, the threshold remains constant. Therefore when the frequency of incident radiation is increased, KE energy will be a little less than double of the initial value.

OR

The Davidson and German experiment confirm the wave nature of electrons.

Mass of electron = m

Charge on electron = e

Let v be the velocity acquired by an electron when accelerated from rest through a potential difference of V volts.

De Broglie's wavelength (λ) = h/mv

For an electron

Work done = gain in it is Kinetic Energy

Work done by an electron = eV

The gain in Kinetic energy =

eV =

2) Mass of electron = m_e

Velocity of an electron = v_e

eV_e =

Mass of proton = m_p

Velocity of proton = V_p

eV_p =

Therefore V_e = V_p

Also, charge on electron = charge on the proton

As for m_e≪ m_p

From

λ_e>λ_p

Therefore De-Broglie wavelength of electron is greater than that of the proton.