If an electron and a proton have the same kinetic energy, the ratio of the de Broglie wavelengths of proton and electron would approximately be:
a) $1:1837$
b) $43:1$
c) $1837:1$
d) $1:43$
Answer
544.2k+ views
Hint: Kinetic energy is the free motion input to a mass to make the identical momentum change or the momentum change that releases the identical amount of free passage of mass onto another mass. Momentum is the result of the product of the mass of the body and its velocity. We will change wavelengths in terms of kinetic energy and mass; then we will get the ratio of wavelengths by putting the mass of electron and proton.
Complete step-by-step solution:
Kinetic energy is given by: $E = \dfrac{1}{2} m v^{2}$
Momentum is given by: $p = mv$
Now we will write momentum in terms of Kinetic energy.
$mv = \sqrt{2mE}$
$\therefore p = \sqrt{2mE}$
Wavelength and momentum are related by:
$\lambda = \dfrac{h}{p}$
$\implies \lambda = \dfrac{h}{\sqrt{2mE}}$
An electron and a proton have the same kinetic energy, E.
Wavelength for electron, $\implies \lambda_{e} = \dfrac{h}{\sqrt{2m_{e}E}}$
Wavelength for proton, $\implies \lambda_{p} = \dfrac{h}{\sqrt{2m_{p}E}}$
Ratio of the de Broglie wavelengths of proton and electron is:
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{h}{\sqrt{2m_{p}E}} \times \dfrac{\sqrt{2m_{e}E}}{h}$
It gives,
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}} $
Now, we have mass of electron, $m_{e} = 9.11 \times 10^{-31} Kg$
We have mass of proton, $m_{p} = 1.67 \times 10^{-27} Kg$
Now, put mass of electron and proton; then we get the ratio of wavelengths.
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{9.11 \times 10^{-31}}{ 1.67 \times 10^{-27}} $
$\implies \dfrac{\lambda_{p} }{\lambda_{e}} = 0.0233$
$\implies \dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{1}{43}$
Option (d) is correct.
Note: The momentum shift and the kinetic energy can be mutually exchangeable; therefore, they can be comparable in quantity and quality. A moving body's kinetic energy and momentum are the body's characteristics that are very much compared to velocity.
Complete step-by-step solution:
Kinetic energy is given by: $E = \dfrac{1}{2} m v^{2}$
Momentum is given by: $p = mv$
Now we will write momentum in terms of Kinetic energy.
$mv = \sqrt{2mE}$
$\therefore p = \sqrt{2mE}$
Wavelength and momentum are related by:
$\lambda = \dfrac{h}{p}$
$\implies \lambda = \dfrac{h}{\sqrt{2mE}}$
An electron and a proton have the same kinetic energy, E.
Wavelength for electron, $\implies \lambda_{e} = \dfrac{h}{\sqrt{2m_{e}E}}$
Wavelength for proton, $\implies \lambda_{p} = \dfrac{h}{\sqrt{2m_{p}E}}$
Ratio of the de Broglie wavelengths of proton and electron is:
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{h}{\sqrt{2m_{p}E}} \times \dfrac{\sqrt{2m_{e}E}}{h}$
It gives,
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}} $
Now, we have mass of electron, $m_{e} = 9.11 \times 10^{-31} Kg$
We have mass of proton, $m_{p} = 1.67 \times 10^{-27} Kg$
Now, put mass of electron and proton; then we get the ratio of wavelengths.
$\dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{9.11 \times 10^{-31}}{ 1.67 \times 10^{-27}} $
$\implies \dfrac{\lambda_{p} }{\lambda_{e}} = 0.0233$
$\implies \dfrac{\lambda_{p} }{\lambda_{e}} = \dfrac{1}{43}$
Option (d) is correct.
Note: The momentum shift and the kinetic energy can be mutually exchangeable; therefore, they can be comparable in quantity and quality. A moving body's kinetic energy and momentum are the body's characteristics that are very much compared to velocity.
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