An electron in the hydrogen atom jumps from n | vedclass

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(C) The energy of the emitted photon $(E)$ is given by the sum of the work function $(\phi)$ and the maximum kinetic energy $(K_{max})$: $E = \phi + K

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$4$

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(C) The energy of the emitted photon $(E)$ is given by the sum of the work function $(\phi)$ and the maximum kinetic energy $(K_{max})$: $E = \phi + K_{max} = 2.65 	ext{ eV} + 10.1 	ext{ eV} = 12.75 	ext{ eV}$. In a hydrogen atom, the energy of an electron in the $n^{th}$ state is $E_n = -13.6 / n^2 	ext{ eV}$. The energy of the photon emitted when jumping from state $n$ to the ground state $(n=1)$ is: $E = E_n - E_1 = -13.6 / n^2 - (-13.6 / 1^2) = 13.6 (1 - 1/n^2) 	ext{ eV}$. Equating the two expressions: $12.75 = 13.6 (1 - 1/n^2)$ $12.75 / 13.6 = 1 - 1/n^2$ $0.9375 = 1 - 1/n^2$ $1/n^2 = 1 - 0.9375 = 0.0625$ $n^2 = 1 / 0.0625 = 16$ $n = 4$.

An electron in the hydrogen atom jumps from $n^{th}$ energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.65 \text{ eV}$. If the maximum kinetic energy of the emitted photoelectrons is $10.1 \text{ eV}$, then the value of '$n$' is

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