Choose the correct statement(s) for hydrogen | vedclass

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(A) When considering the motion of the nucleus,the mass of the electron $m_e$ is replaced by the reduced mass $\mu = \frac{m_e M}{m_e + M}$,where $M$

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The radius of the first Bohr orbit of deuterium is less than that of hydrogen.

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(A) When considering the motion of the nucleus,the mass of the electron $m_e$ is replaced by the reduced mass $\mu = \frac{m_e M}{m_e + M}$,where $M$ is the mass of the nucleus.Since the mass of the deuterium nucleus $(M_D \approx 2M_H)$ is greater than that of the hydrogen nucleus,the reduced mass $\mu_D$ is greater than $\mu_H$.The radius of the $n^{th}$ Bohr orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi \mu e^2}$. Since $\mu_D > \mu_H$,the radius $r_D The speed of the electron is given by $v_n = \frac{e^2}{2 n h \epsilon_0}$. This expression does not depend on the mass of the nucleus,so $v_D = v_H$. Option $B$ is incorrect.The wavelength of the emitted photon is given by $\frac{1}{\lambda} = R_{\mu} Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$,where $R_{\mu} = \frac{\mu e^4}{8 \epsilon_0^2 h^3 c}$. Since $\mu_D > \mu_H$,$R_D > R_H$. Consequently,$\lambda_D The angular momentum of the electron in the $n^{th}$ orbit is $L = n \frac{h}{2\pi}$,which is independent of the mass of the nucleus. Thus,$L_D = L_H$. Option $D$ is incorrect.

Choose the correct statement$(s)$ for hydrogen and deuterium atoms (considering the motion of the nucleus).

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