Ncert Exercises & Solutions

Question 11.23:

(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

(a)
Hint: $E=\frac{hc}{\lambda }$
Step 1: Find The maximum energy of a photon.
Wavelength produced by an X-ray tube,

λ

= 0.45 Å = 0.45 x 10^-10 m

Planck's constant, h = 6.626 x 10^-34 Js
Speed of light, c = 3 x 10⁸ m/s
The maximum energy of a photon is given as:

$\begin{aligned} E & = \frac{hc}{λ} \\ = \frac{6 .626 \times 10^{- 34} \times 3 \times 10^{8}}{0 .45 \times 10^{- 10} \times 1 .6 \times 10^{-19}} \\ = 27 .6 \times 10^{3} eV = 27 .6 keV \end{aligned}$

Therefore, the maximum energy of an X-ray photon is 27.6 keV.

(b)
Hint: Accelerating voltage provides energy to the electrons for producing X-rays.
Step 1: Find the order of accelerating voltage.
To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions