A hydrogen atom in the ground state is excite | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \; 	ext{eV} \cdot \mathring{A}$,we get $E = \frac{

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$6$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \; 	ext{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{975} \approx 12.75 \; 	ext{eV}$.The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \; 	ext{eV}$.For the ground state $(n=1)$,$E_1 = -13.6 \; 	ext{eV}$.After absorbing the photon,the new energy level is $E_n = E_1 + E = -13.6 + 12.75 = -0.85 \; 	ext{eV}$.Since $E_n = -\frac{13.6}{n^2} = -0.85 \; 	ext{eV}$,we have $n^2 = \frac{13.6}{0.85} = 16$,so $n = 4$.The electron is excited to the $n = 4$ state.The number of spectral lines emitted when the electron transitions from $n$ to the ground state is given by $\frac{n(n-1)}{2}$.For $n = 4$,the number of spectral lines is $\frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$.

$A$ hydrogen atom in the ground state is excited by monochromatic radiation of $\lambda = 975 \; \mathring{A}$. The number of spectral lines in the resulting emission spectrum will be:

Similar Questions

$A$ hydrogen atom transitions from an excited state to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant,the principal quantum number $n$ of the excited state is:

In terms of Rydberg constant $R,$ the shortest wavelength in the Balmer series of the Hydrogen atom spectrum will be:

Which of the following statements is true?

$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

What is the shortest wavelength present in the Paschen series of spectral lines (in $nm$)?

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