Question

An electron in a hydrogen like atom makes transition from a state in which its de-Broglie wavelength is l1 to a state where its de-Broglie wavelength is l2 then wavelength of photon (l) generated will be

• A. $\lambda = \lambda_{1} - \lambda_{2}$
• B. $\lambda = \frac{4mc}{h}\left\{ \frac{\lambda_{1}\lambda_{2}}{\lambda_{1}^{2} - \lambda_{2}^{2}} \right\}$
• C. $\lambda = \sqrt{\frac{\lambda_{1}\lambda_{2}}{\lambda_{2}^{2} - \lambda_{2}^{2}}}$
• D. $\lambda = \frac{2mc}{h}\left\{ \frac{\lambda_{1}\lambda_{2}}{\lambda_{1} - \lambda_{2}^{2}} \right\}$

Answer 1

Answer: (D)

$\lambda = \frac{2mc}{h}\left\{ \frac{\lambda_{1}\lambda_{2}}{\lambda_{1} - \lambda_{2}^{2}} \right\}$

Answer 2

$\frac{hc}{\lambda} = E_{1} - E_{2} = KE_{2} - KE_{1}$

$\therefore\lambda = \frac{h}{mV} (mV)^{2} = \left( \frac{h}{\lambda} \right)^{2}; \frac{1}{2}mV^{2} = \frac{1}{2m}\frac{h^{2}}{\lambda^{2}}\therefore$ $\frac{hc}{\lambda} = \frac{h^{2}}{2m\lambda_{2}^{2}} - \frac{h^{2}}{2m\lambda_{1}^{2}} \therefore \lambda = \frac{2mc}{h}\left\{ \frac{\lambda_{1}^{2}\lambda_{2}^{2}}{\lambda_{1}^{2} - \lambda_{2}^{2}} \right\}$