Question

Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Answer 1

Given in the question :-

Now for the maximum frequency :-

$f_m_a_x$ = eV /h

$f_m_a_x$  = (1.6 × 10⁻¹⁹ × 30× 10³) / 6.626 × 10⁻³⁴

$f_m_a_x$  = 7.244 × 10¹⁸ Hz

Now for the second part :- b

Minimum wavelength of x-rays

$\lambda_m_i_n$ = c / $f_m_a_x$

$\lambda_m_i_n$ = 3 × 10⁸ / 7.244 × 10¹⁸

4.141 × 10⁻¹¹ m

$\lambda_m_i_n$  = 0.4141 A°

Hope it helps :-)

Answer 2

★ Given:-

Electron potential, V = 30 kV = 3 × 10⁴ V

Electron energy, E = 3 × 10⁴ eV

• Where,

e = Charge on one electron = $1.6\ times{10}^{-19}C$

(a) Maximum frequency by the X-rays = ν

★ The energy of the electrons → E = hν

• Where,

h = Planck’s constant = $6.626\times {10}^{-34}Js$

→ Therefore, $v = \dfrac{E}{h}$

$= \frac{1.6 \times {10}^{ - 19} \times 3 \times {10}^{4}}{6.626 \times {10}^{ - 34} } \\ \\ = 7.24 \times {10}^{18}Hz$

→ Hence, $7.24\times {10}^{18} Hz$ is the maximum frequency of the X-rays.

(b) The minimum wavelength produced:

$\lambda =\dfrac{c}{v}$

$= \dfrac{3 \times {10}^{8} }{7.24 \times {10}^{18}} \\ \\ = 4.14 \times {10}^{ - 11} m$

$= 0.0414 \: nm$