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
585.3k+ 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.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

