Question

When the kinetic energy of an electron is increased, the wavelength of the associated wave will:
(a) increase
(b) decrease
(c) remain independent of kinetic energy
(d) be none of the above

Answer 1

Hint: Wavelength is inversely proportional to square root of kinetic energy.

Formula used:
> Wavelength:
$\lambda ={\displaystyle \frac{h}{p}}$ ……(1)
where,
h is the Planck’s constant
p is the momentum
> Momentum:
$p={\displaystyle \frac{h}{\sqrt{2mE}}}$ ……(2)
where,
m is the mass
E is the kinetic energy

Complete step-by-step answer:
Given:
1. Kinetic energy is increased.

To find: The new wavelength of the associated wave.

Step 1 of 2:
Write eq (1) by substituting the expression for p:
$\lambda ={\displaystyle \frac{h}{\sqrt{2mE}}}$ ……(3)

We can see that:
$\lambda \propto {\displaystyle \frac{1}{\sqrt{E}}}$

Step 2 of 2:
Wavelength and kinetic energy are inversely proportional to each other. Increasing one, will decrease the other and vice versa.
Hence, if we increase the kinetic energy, the wavelength will decrease.

Correct answer:
When the kinetic energy of an electron is increased, the wavelength of the associated wave will: (b) decrease.

Additional Information:
Higher the energy lowers the wavelength. As higher the energy higher will be frequency. But frequency varies inversely with wavelength. Therefore, lower would be wavelength. De-Broglie wavelength associated will also decrease. In the same trend as above. A wave can be superposition of multiple wavelengths at a time but not for the same particle in a definite trajectory with definite velocity.

Note: In questions like these, remember the formula for wavelength and the relation between wavelength and kinetic energy. If two quantities are directly proportional, increasing one will increase the other. On the other hand, if two quantities are inversely proportional, increasing one will decrease the other.