Question

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.

1. `E_e/E_p = 10^-4`
2. `E_e/E_p = 10^-2`
3. `p_e/(m_ec) = 10^-2`
4. `p_e/(m_ec) = 10^-4`

Options

• b and c

• a and c

• c and d

• b and d

Answer 1

b and c

Explanation:

De-Broglie wavelength associated with the charged particles: The energy of a charged particle accelerated through potential difference V is `E = 1/2 mv^2 = qV`

Hence de-Broglie wavelength `λ = h/p = h/sqrt(2mE) = h/sqrt(2mqV)`

`λ_("Electron") = 12.27/sqrt(V) Å, λ_("Proton") = 0.286/sqrt(V) Å`

`λ_("Deutron") = 0.202/sqrt(V) Å, λ_(α"-particle") = 0.101/sqrt(V) Å`

De-Broglie wavelength associated with uncharged particles: For Neutron de-Broglie wavelength is given as

`λ_("Neutron") = (0.286 xx 10^-10)/sqrt(E ("in"  e V)) m = 0.286/sqrt(E("in"  eV)) Å`

Energy of thermal neutrons at ordinary temperature

∵ E = kT ⇒ λ = `h/sqrt(2mkT)`; where T = Absolute temperature,

k = Boltzman's constant = 1.38 × 10^–23 Joule/kelvin

So, `λ_("Thermal neutron") = (6.62 xx 10^-34)/sqrt(2 xx 1.67 xx 10^-27 xx 1.38 xx 10-23 T) = 30.83/sqrt(T) Å`

Mass of electron = m_e, Mass of photon = m_p,

Velocity of electron is equal to velocity of proton which are v_e and v_p respectively.

De-Broglie wavelength for electron, `λ_e h/sqrt(m_e v_e)`

⇒ `λ_e = h/(m_e(c/100)) = (100 h)/(m_e c)`  .....(i)

Kinetic energy of electron, `E_e = 1/2 m_e v_e^2`

⇒ `m_e v_e = sqrt(2E_e m_e)`

So, `λ_e = h/(m_e v_e) = h/sqrt(2m_e E_e)`

⇒ `E_e = h^2/(2λ_e^2 m_e)`  .....(ii)

The wavelength of proton is λ_p and having an energy is

∴ `E_p = (hc)/λ_p = (hc)/(2λ_e)`   ......[∵ λ_p = 2λ_e]

∴ `E_p/E_e = ((hc)/(2λ_e)) ((2λ_e^2 m_e)/h^2)`

= `(λ_em_ec)/h = (100h)/(m_ec) xx (m_ec)/h` = 100

So, `E_e/E_p = 1/100 = 10^-2`

For electron, `p_e = m_ev_e = m_e xx c/100`

So, `p_e/(m_ec) = 1/100 = 10^-2`

Important point:

Ratio of the wavelength of photon and electron: The wavelength of a photon of energy E is given by `λ_(ph) = (hc)/E`

While the wavelength of an electron of kinetic energy K is given by `λ_c = h/sqrt(2mK)`

Therefore for the same energy the ratio `λ_(ph)/λ_e = c/E sqrt(2mK) = sqrt((2mc^2 K)/E^2`