An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :

Given, V = 1.24 million volt = 1.24 $\times$ 10⁶ volt<br/><br/>Since, energy (E) = eV<br/><br/>where, e is the charge of electron = 1.6 $\times$ 10^$-$19 C<br/><br/>$\therefore$ E = 1.6 $\times$ 10^$-$19 $\times$ 1.24 $\times$ 10⁶ ..... (i)<br/><br/>As we know that,<br/><br/>Energy of photon, $E = {{hc} \over \lambda }$ .... (ii)<br/><br/>Here, Planck's constant, h = 6.67 $\times$ 10^$-$34 J-s,<br/><br/>c = speed of light in free space, c = 3 $\times$ 10⁸ ms^$-$1<br/><br/>Equating Eqs. (i) and (ii), we get<br/><br/>$$1.6 \times {10^{ - 19}} \times 1.24 \times {10^6} = {{6.67 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over \lambda }$$<br/><br/>$$ \Rightarrow \lambda = {{20.01 \times {{10}^{ - 13}}} \over {1.6 \times 1.24}} = 10.09 \times {10^{ - 13}}$$<br/><br/>$= 1.009 \times {10^{ - 12}} \simeq {10^{ - 3}} \times {10^{ - 9}}$<br/><br/>= 10^$-$3 nm