Question

The de-Broglie wavelength of a proton accelerated by $ 400V $ is?

Answer 1

Hint: The potential of a proton is given as $ 400V $ . The de-Broglie wavelength of a proton can be determined from the below formula, for this the kinetic energy increases will be needed which is equal to the work done by the electric field. The mass and charge of the proton were also needed in the calculation of wavelength.
 $ \lambda = \dfrac{h}{{mv}} $
 $ \lambda $ is de-Broglie wavelength
 $ h $ is Planck’s constant
 $ m $ is mass of proton
 $ v $ is velocity of proton
 $ v = \sqrt {\dfrac{{2qV}}{m}} $
 $ q $ is charge of proton
 $ V $ is the potential of protons.

Complete answer:
The de-Broglie wavelength has the formula of $ \lambda = \dfrac{h}{{mv}} $ in which the velocity will be determined from the kinetic energy equation.
The work done to increase the potential will be equal to the kinetic energy of a proton.
The kinetic energy is given as $ \dfrac{1}{2}m{v^2} $ and the increase in potential will be $ qV $ , by equating these both equations, $ qV = \dfrac{1}{2}m{v^2} $
The velocity can be written as $ v = \sqrt {\dfrac{{2qV}}{m}} $
The potential of an electron is given as $ 400V $
Substitute the charge of proton as $ 1.6 \times {10^{ - 19}} $ and potential as $ 400V $ , the mass of protons is $ 1.6 \times {10^{ - 27}} $
Thus, velocity is $ v = \sqrt {\dfrac{{2 \times 1.6 \times {{10}^{ - 19}} \times 400}}{{1.6 \times {{10}^{ - 27}}}}} = 2.77 \times {10^5}m{\left( {\sec } \right)^{ - 1}} $
Now, substitute this velocity in the de-Broglie wavelength
Substitute the value of Planck’s constant as $ 6.6 \times {10^{ - 34}} $
 $ \lambda = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{\left( {1.6 \times {{10}^{ - 27}}} \right)\left( {2.77 \times {{10}^5}} \right)}} = 1.43 \times {10^{ - 12}}m $
Thus, the de-Broglie wavelength of a proton accelerated by $ 400V $ is $ 1.43 \times {10^{ - 12}}m $

Note:
The work done to increase the electric field is the product of the charge and potential of a proton. It should be equated with the kinetic energy of a proton. From these equations only the velocity can be determined. The velocity must be in metres per second to obtain the wavelength in metres.