Light from the Paschen series of a hydrogen a | vedclass

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(D) In the Paschen series,an electron transitions from a higher energy state to the $n=3$ state. The energy of a photon in the Paschen series is given

Vedclass · Accepted Answer

$1.1 	ext{ eV}$

Vedclass · Answer

(D) In the Paschen series,an electron transitions from a higher energy state to the $n=3$ state. The energy of a photon in the Paschen series is given by $E = 13.6 \left( \frac{1}{3^2} - \frac{1}{n^2} ight) 	ext{ eV}$,where $n = 4, 5, 6, \dots$The minimum energy corresponds to the transition from $n=4$ to $n=3$:$E_{\min} = 13.6 \left( \frac{1}{9} - \frac{1}{16} ight) = 13.6 \left( \frac{16-9}{144} ight) = 13.6 \left( \frac{7}{144} ight) \approx 0.66 	ext{ eV}$.The maximum energy corresponds to the transition from $n=\infty$ to $n=3$:$E_{\max} = 13.6 \left( \frac{1}{9} - 0 ight) = \frac{13.6}{9} \approx 1.51 	ext{ eV}$.For a photon to eject a photoelectron,its energy must be greater than or equal to the work function $\phi_0$ of the metal. Thus,$\phi_0 \leq E_{	ext{photon}}$. Since the maximum energy of a Paschen photon is $1.51 	ext{ eV}$,the work function must satisfy $\phi_0 \leq 1.51 	ext{ eV}$.Among the given options,$1.1 	ext{ eV}$ is the only value that satisfies this condition.

Light from the Paschen series of a hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is:

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