The K_{alpha} X-ray emission from tungsten oc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy difference between the $K$ and $L$ levels is given by the energy of the emitted $K_{\alpha}$ photon: $E = \frac{hc}{\lambda}$ Substitut

Vedclass · Accepted Answer

$59 \ keV$

Vedclass · Answer

(C) The energy difference between the $K$ and $L$ levels is given by the energy of the emitted $K_{\alpha}$ photon: $E = \frac{hc}{\lambda}$ Substituting the values: $E = \frac{(6.63 	imes 10^{-34} \ J \cdot s) 	imes (3 	imes 10^8 \ m/s)}{0.021 	imes 10^{-9} \ m}$ $E \approx 9.47 	imes 10^{-15} \ J$ To convert this energy into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \ J/eV$: $E = \frac{9.47 	imes 10^{-15}}{1.6 	imes 10^{-19}} \ eV \approx 59,187 \ eV$ $E \approx 59 \ keV$.

The $K_{\alpha}$ $X$-ray emission from tungsten occurs at $\lambda = 0.021 \ nm$. The energy difference between the $K$ and $L$ levels in the atom is:

Similar Questions

The minimum wavelength of $X$-rays produced by $20 \ kV$ electrons is nearly (in $Å$)

$X-$ray beams can be deflected by

$X$-rays of which of the following wavelengths are hardest? (in $\mathring{A}$)

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on a target. The cut-off wavelength $\lambda_0$ is equal to [$h=$ Planck's constant,$c=$ velocity of light].

The wavelength of $X$-rays is of the order of ...... $ \mathring A $.

Vedclass Test Series

Exam Paper Generator

Online Exam Module