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Question: An electron has a wavelength of \[1\,{\text{nm}}\] . Then, the kinetic energy of the electron is: ...

An electron has a wavelength of 1 nm1\,{\text{nm}} . Then, the kinetic energy of the electron is:
(A) 1.8×10−6 J1.8 \times {10^{ - 6}}\,{\text{J}}
(B) 19.86×10−17 J19.86 \times {10^{ - 17}}\,{\text{J}}
(C) 16.82×10−19 J16.82 \times {10^{ - 19}}\,{\text{J}}
(D) 22.3×10−9 J22.3 \times {10^{ - 9}}\,{\text{J}}

Explanation

Solution

First of all, we will use the relation which relates energy and the wavelength, basically called the Planck’s equation. We will substitute the required values and manipulate accordingly to obtain the answer.

Complete step by step solution:
We know that a moving electron possesses both the particle behaviour along with the wave behaviour. In order to find the kinetic energy associated with the electron, we can simply use Planck's equation which gives a relation between the energy and the frequency of the subatomic particles. The relation can also be modified as the relation between the energy and the wavelength.
Let us write the Planck’s equation, which is given below:
E=hνE = h\nu …… (1)
Where,
EE indicates kinetic energy.
hh indicates Planck’s constant.
ν\nu indicates the frequency of the electron.
As we know, the relation between the frequency and the wavelength as:
ν=cλ\nu = \dfrac{c}{\lambda } …… (2)
Where,
ν\nu indicates the frequency of the electron.
cc indicates the speed of light rays containing the electron.
λ\lambda indicates the wavelength of the moving electron.
Again, we can modify the equation (1) by using the equation (2), and we get:
E=h×cλE = h \times \dfrac{c}{\lambda } …… (3)
Now, we substitute the required values in the equation (3) and we get:
E=h×cλ ⇒E=6.626×10−34×3×1081×10−9 ⇒E=1.986×10−16J ∴E=19.86×10−17JE = h \times \dfrac{c}{\lambda } \\\ \Rightarrow E = 6.626 \times {10^{ - 34}} \times \dfrac{{3 \times {{10}^8}}}{{1 \times {{10}^{ - 9}}}} \\\ \Rightarrow E = 1.986 \times {10^{ - 16}}{\text{J}} \\\ \therefore E = 19.86 \times {10^{ - 17}}{\text{J}}
Hence, its kinetic energy in terms of joules is 19.86×10−17J19.86 \times {10^{ - 17}}{\text{J}} .

The correct option is B.

Note: While solving this problem, remember that as the electron is moving, it possesses kinetic energy by virtue of its mass and velocity. Although its mass is very small, the velocity at which it is travelling is really high. It is an advice to take all the units in the S.I system of units. If not done so, we may get irrelevant results. Although the energy we have got is very small, but in the large scale it really counts.