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(D) The energy difference between $n = 2$ and $n = 1$ levels in a hydrogen atom is $\Delta E = 13.6 \ eV 	imes (1 - 1/4) = 10.2 \ eV$.When the atom e

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Kinetic energy of recoil $H$-atom will be equal to energy of emitted photon.

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(D) The energy difference between $n = 2$ and $n = 1$ levels in a hydrogen atom is $\Delta E = 13.6 \ eV 	imes (1 - 1/4) = 10.2 \ eV$.When the atom emits a photon, it must recoil to conserve momentum. Let $p_p$ be the momentum of the photon and $p_a$ be the momentum of the atom. By conservation of momentum, $p_a = p_p = E/c$.The total energy released is $10.2 \ eV$, which is shared between the photon energy $(E_p)$ and the recoil kinetic energy of the atom $(K_a)$. Thus, $E_p + K_a = 10.2 \ eV$.Since $K_a > 0$, $E_p Since $K_a = 10.2 \ eV - E_p$ and $E_p$ is very close to $10.2 \ eV$, $K_a$ is very small, so $K_a Statement $(d)$ is incorrect because the total energy $10.2 \ eV$ is shared; the kinetic energy of the atom is much smaller than the photon energy.Statement $(a)$ is incorrect because the De-Broglie wavelength $\lambda_a = h/p_a$ and photon wavelength $\lambda_p = hc/E_p$. Since $p_a = E_p/c$, we have $\lambda_a = hc/E_p = \lambda_p$. Wait, actually, $p_a = p_p$, so $\lambda_a = h/p_a = h/p_p = \lambda_p$. Thus, $(a)$ is technically correct. Given the options, $(d)$ is the most clearly incorrect statement.

$A$ hydrogen atom at rest emits a photon during its transition from $n = 2$ to $n = 1$. Choose the $INCORRECT$ statement.

Similar Questions

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The speed of an electron in the orbit of a hydrogen atom in the ground state is

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