Question

The energy of a K-electron in tungsten is $-20keV$ and of an L-electrons is $-2keV$. The wave length of X-rays emitted when there is electron jump from L to K shell :-

• A. $0.3443Ă$
• B. $0.6887Ă$
• C. $1.3982Ă$
• D. $2.78Ă$

Answer 1

The correct option is B

$0.6887Ă$

Step - 1: Given

Energy of K-electron in tungsten $E_K=-20keV=-20\times {10}^3\times 1.69\times {10}^{-19}J$

Energy of L-electron in tungsten $E_L=-2keV=-2\times {10}^3\times 1.69\times {10}^{-19}J$

Step - 2: Formula

The energy of an electron is given by, $E=h\nu =\frac{hc}{\lambda }$, where h is Planck's constant $6.62\times {10}^{-34}J/Hz$and c is Speed of ray $3\times {10}^{8}m/s$ and $\lambda$is the wavelength

Step - 3: Wavelength of X-ray emitted

$$\begin{gathered} E=E_K-E_L \\ \Rightarrow E_K-E_L=\frac{hc}{\lambda } \\ \Rightarrow \lambda =\frac{hc}{E_K-E_L} \\ \Rightarrow \lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{\left(-20\times {10}^3\times 1.69\times {10}^{-19}\right)-\left(-2\times {10}^3\times 1.69\times {10}^{-19}\right)}m \\ \Rightarrow \lambda =0.6887Ă \end{gathered}$$

Hence option B is correct.