Correct Answer - Option 1 : 6.62 × 10
-13 m
Concept:
De Broglie proposed that the wavelength λ associated with a particle of momentum p is given as:
\(\lambda = \frac{h}{p} = \frac{h}{{mv}}\)
Where m is the mass of the particle and v its speed. Planck’s Constant, h = 6.623 × 10-34 Js
The above equation is known as the de Broglie relation and the wavelength λ of the matter-wave is called de Broglie wavelength.
Thus, the significance of the de-Broglie equation lies in the fact that it relates the particle character to the wave character of matter.
Note:
According to Planck’s quantum theory, the energy of a photon is described as:
E = hν
According to Einstein’s mass-energy relation:
E = mc2
Frequency ν can be expressed in terms of wavelength λ as:
ν = c/λ
hν = mc2
hν/c = mc
λ = h/mc
This equation is applicable for a photon.
Calculation:
\(\lambda = \frac{h}{{mv}} = \frac{{6.623 \times {{10}^{ - 34}}}}{{1.67 \times {{10}^{ - 27}} \times 6 \times {{10}^5}}} \)
\(= 0.661 \times {10^{ - 12}}\;m \)
\(= 6.61 \times {10^{ - 13}}\;m\)
