• # question_answer If${{\lambda }_{1}}$and${{\lambda }_{2}}$denote the wavelengths of de Broglie waves for electrons in Bohr's first and second orbits in the hydrogen atom, then${{\lambda }_{1}}/{{\lambda }_{2}}$will be: A)  2                                B) $\frac{1}{2}$C)  4                                D) $\frac{1}{4}$

De-Broglie wavelength $=\lambda =\frac{h}{\rho }$ $\therefore$$\lambda =\frac{h}{\sqrt{2mk}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{k}}$ Where, k is the kinetic energy of the electron. Now $\lambda =\frac{h}{\sqrt{2mk}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{k}}$ Hence, the correction option is [b].