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प्रश्न
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 ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
पर्याय
b and c
a and c
c and d
b and d
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उत्तर
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 = me, Mass of photon = mp,
Velocity of electron is equal to velocity of proton which are ve and vp 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 λp and having an energy is
∴ `E_p = (hc)/λ_p = (hc)/(2λ_e)` ......[∵ λp = 2λe]
∴ `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`
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