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Question: Calculate the accelerating potential that must be imparted to a proton is to give it an effective wa...

Calculate the accelerating potential that must be imparted to a proton is to give it an effective wavelength of 0.005 nm{\text{0}}{\text{.005 nm}} .
A. V = 13.6 volt{\text{V = 13}}{\text{.6 volt}}
B. V = 32.82 volt{\text{V = 32}}{\text{.82 volt}}
C. V = 10.4 volt{\text{V = 10}}{\text{.4 volt}}
D. V = 1.23 volt{\text{V = 1}}{\text{.23 volt}}

Explanation

Solution

First calculate the velocity of the proton by using the De-Broglie hypothesis. Then equate the kinetic energy of the proton with its potential energy.

Complete answer:
According to De-Broglie hypothesis,
λ = hm×u\lambda {\text{ }} = {\text{ }}\dfrac{h}{{m \times u}}
Here, λ\lambda is the wavelength of a proton of mass m moving with a velocity of u. h is Planck’s constant.
Rearrange above equation:
⇒u = hm×λ\Rightarrow u{\text{ }} = {\text{ }}\dfrac{h}{{m \times \lambda }} … …(1)
You can use equation (1) to calculate the velocity of the proton.
In the equation (1), substitute 6.626×10−34Js6.626 \times {10^{ - 34}}Js for planck’s constant, 1.672×10−27kg1.672 \times {10^{ - 27}}kg for mass of proton, 0.005 nm{\text{0}}{\text{.005 nm}} for the wavelength of proton and use the conversion factor 10−9mnm\dfrac{{{{10}^{ - 9}}m}}{{nm}} to convert unit of wavelength from nanometer to meter.

⇒u = hm×λ ⇒u = 6.626×10−34Js1.672×10−27kg×(0.005 nm×10−9mnm)4 ⇒u=7.93×104meter sec−1\Rightarrow u{\text{ }} = {\text{ }}\dfrac{h}{{m \times \lambda }} \\\ \Rightarrow u{\text{ }} = {\text{ }}\dfrac{{6.626 \times {{10}^{ - 34}}Js}}{{1.672 \times {{10}^{ - 27}}kg \times \left( {{\text{0}}{\text{.005 nm}} \times \dfrac{{{{10}^{ - 9}}m}}{{nm}}} \right)}}4 \\\ \Rightarrow u = 7.93 \times {10^4}{\text{meter se}}{{\text{c}}^{ - 1}}

Hence, the proton is moving at a velocity of 7.93×104meter sec−17.93 \times {10^4}{\text{meter se}}{{\text{c}}^{ - 1}} .
Let V be the accelerating potential.
A proton having charge Q and mass m will acquire the energy E=QVE = QV
But this energy is equal to the kinetic energyE=12mu2E = \dfrac{1}{2}m{u^2}
Hence,
QV=12mu2QV = \dfrac{1}{2}m{u^2}
Rearrange above expression
⇒u=2QVm\Rightarrow u = \sqrt {\dfrac{{2QV}}{m}}
Substitute 1.602×10−191.602 \times {10^{ - 19}} for Q, and 1.672×10−271.672 \times {10^{ - 27}} for the mass of proton in the above expression and get an expression in terms of u.

⇒u=2×1.602×10−19×V1.672×10−27 ⇒u=1.916×108×V ⇒u=1.384×104×V\Rightarrow u = \sqrt {\dfrac{{2 \times 1.602 \times {{10}^{ - 19}} \times V}}{{1.672 \times {{10}^{ - 27}}}}} \\\ \Rightarrow u = \sqrt {1.916 \times {{10}^8} \times V} \\\ \Rightarrow u = 1.384 \times {10^4} \times \sqrt V

But, from De-Broglie hypothesis
⇒u=7.93×104meter sec−1\Rightarrow u = 7.93 \times {10^4}{\text{meter se}}{{\text{c}}^{ - 1}}
Hence,

⇒1.384×104×V=7.93×104meter sec−1 ⇒V=7.93×1041.384×104 ⇒V=5.728\Rightarrow 1.384 \times {10^4} \times \sqrt V = 7.93 \times {10^4}{\text{meter se}}{{\text{c}}^{ - 1}} \\\ \Rightarrow \sqrt V = \dfrac{{7.93 \times {{10}^4}}}{{1.384 \times {{10}^4}}} \\\ \Rightarrow \sqrt V = 5.728

Take square on both sides of equation
⇒V = 32.82 volt\Rightarrow {\text{V = 32}}{\text{.82 volt}}
Hence, the accelerating potential of 32.82 volt{\text{32}}{\text{.82 volt}} must be imparted to a proton to give it an effective wavelength of 0.005 nm{\text{0}}{\text{.005 nm}} .

**Hence, the correct option is the option B.

Note:**
De-Broglie hypothesis explains the dual behavior of matter. Matter can also behave as a wave and at the same time, waves can show particle nature.