Advertisements
Advertisements
Question
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Advertisements
Solution 1
The momentum of a photon having energy (hv) is given as:
p = `(h v)/c`
= `h/lambda`
λ = `h/p` ...(i)
Where,
λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:
λ = `h/(m v)`
But p = mv
∴ λ = `h/p` ...(ii)
Where,
m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Solution 2
For a photon of electromagnetic radiation: From Planck’s theory:
E = `(h c//lambda_(em))/c`
= `h/(lambda_(em))`
Rearranging for wavelength:
λem = `h/p` ...(i)
According to de Broglie, the wavelength of any “quantum” (particle) is: λdB = `h/p` ...(ii)
Comparing equations (i) and (ii), we get:
λem = λdB
RELATED QUESTIONS
Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of `(3/2)` kT at 300 K.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10−31 kg).
The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon in (2λmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
When a light wave travels from air to glass ______.
Show with the help of a labelled graph how their wavelength (λ) varies with their linear momentum (p).
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
70 cal of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the gas through the same range at constant volume will be (assume R = 2 cal/mol-K).
A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.
An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then ______
An electron is moving with an initial velocity `v = v_0hati` and is in a magnetic field `B = B_0hatj`. Then it’s de Broglie wavelength ______.
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`
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
An electron is accelerated from rest through a potential difference of 100 V. Find:
- the wavelength associated with
- the momentum and
- the velocity required by the electron.
The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen is ______ × 10-11 m (given radius of first Bohr orbit is 5.3 × 10-11 m):
How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
