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Question: If the deBroglie wavelength of an electron is equal to \({10^{ - 3}}\) times the wavelength of a pho...

If the deBroglie wavelength of an electron is equal to 10−3{10^{ - 3}} times the wavelength of a photon of frequency 6×10146 \times {10^{14}} Hz, then what is the speed of electron? (Speed of light = 3×1083 \times {10^8} m/s, Planck’s constant = 6.63×10−346.63 \times {10^{ - 34}} J, Mass of electron = 9.11×10−319.11 \times {10^{ - 31}} kg)
(A) 1.45×1061.45 \times {10^6} m/s
(B) 1.75×1061.75 \times {10^6} m/s
(C) 1.8×1061.8 \times {10^6} m/s
(D) 1.1×1061.1 \times {10^6} m/s

Explanation

Solution

The wavelength of the photon can be calculated from the frequency given, which can be used to find the wavelength of the electron. The speed of the electron and its wavelength are related by the expression for deBroglie wavelength.

Complete step by step answer:
The deBroglie wavelength of the electron is given to be 1000 times the wavelength of a photon of frequency 6×10146 \times {10^{14}} Hz. First, we need to find the wavelength of the photon in order to find out the wavelength of the electron. We know the frequency of the photon. The wavelength of the photon can be found out by using,
fp=cλp{f_p} = \dfrac{c}{{{\lambda _p}}}
⇒λp=cfp=3×1086×1014=5×10−6\Rightarrow {\lambda _p} = \dfrac{c}{{{f_p}}} = \dfrac{{3 \times {{10}^8}}}{{6 \times {{10}^{14}}}} = 5 \times {10^{ - 6}} m
Now that we have found out the wavelength of the photon, we can find out the wavelength of the electron, which is 1000 times the wavelength of the photon.
λ=10−3λp=5×10−9\lambda = {10^{ - 3}}{\lambda _p} = 5 \times {10^{ - 9}} m
The wavelength and the velocity of the electron are related by the expression for the deBroglie wavelength of the electron. The expression is,
hp=λ\dfrac{h}{p} = \lambda
p is the momentum of the electron. It is equal to the product of the mass of the electron and the velocity of the electron.
hmv=λ ⇒v=hmλ\dfrac{h}{{mv}} = \lambda \\\ \Rightarrow v = \dfrac{h}{{m\lambda }} …equation (1)
On substituting the values of planck’s constant, mass of electron and the wavelength of the electron in equation (1), we get,
∴v=6.63×10−349.11×10−31×5×10−9=1.45×106\therefore v = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 5 \times {{10}^{ - 9}}}} = 1.45 \times {10^6} m/s

Hence, option A is the correct answer.

Note: According to deBroglie, matter also exhibits the properties of waves, hence the relations valid for a photon are also valid for particles. Therefore, energies of the photon and particles can also be equated in such a question.