Question

When an electron is accelerated through a 480V, the wavelength is $ \lambda $. Find the wavelength in terms of $ \lambda $ if it is accelerated through 120V.

• A. \( \lambda/2 \)
• B. \( 2\lambda \)
• C. \( \lambda/4 \)
• D. \( \lambda/3 \)

Hint:

When an electron is accelerated through a higher voltage, its wavelength decreases, and the relationship is inverse, so lowering the voltage increases the wavelength.

Answer 1

The Correct Option is A

When an electron is accelerated through a voltage \( V \), the wavelength \( \lambda \) of the emitted radiation is related to the energy \( E \) acquired by the electron. The energy acquired is given by: \[ E = eV \] where \( e \) is the charge of the electron and \( V \) is the voltage. The energy of the electron is inversely proportional to the wavelength, as given by: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. From this, we can derive that the wavelength is inversely proportional to the accelerating voltage: \[ \lambda \propto \frac{1}{V} \] Given that the initial wavelength for a voltage of 480V is \( \lambda \), for a voltage of 120V, the wavelength is: \[ \lambda' = \frac{\lambda}{\frac{480}{120}} = \frac{\lambda}{4} \] 
Thus, the new wavelength when the electron is accelerated through 120V is \( \frac{\lambda}{4} \).