The wavelength $λ_e$ of an electron and $λ_p$ of a photon of same energy E are related by

Question

In a Vernier caliper, $N+1$ divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:

Answer 1

To relate the wavelength of an electron λ_e and a photon λ_p when both have the same energy E, we start by considering the de Broglie wavelength and the photon wavelength formulae:

Wavelength of an electron:

The de Broglie wavelength λ_e of an electron is given by:

λ_e = \frac{h}{\sqrt{2mE}}

where:
- h is Planck's constant
- m is the mass of the electron
- E is the energy
Wavelength of a photon:

The wavelength λ_p of a photon is given by:

λ_p = \frac{hc}{E}

where:
- c is the speed of light
- E is the energy

Now, we need to express λ_p in terms of λ_e:

Starting from the expression for λ_e,

λ_e = \frac{h}{\sqrt{2mE}} \Rightarrow \sqrt{2mE} = \frac{h}{λ_e}

Therefore, the energy E can be expressed as:

E = \frac{h^2}{2mλ_e^2}

Inserting this expression for E into the formula for λ_p gives:

λ_p = \frac{hc}{E} = \frac{hc}{\frac{h^2}{2mλ_e^2}} = \frac{2mcλ_e^2}{h}

Thus, λ_p ∝ λ_e^2.

The correct answer is λ_p ∝ λ_e^2, which corresponds to the given correct option.

Answer 2