Hydrogen atom in its ground state is excited by radiation of wavelength 975 A- How many lines will be there in the emission spectrumA- 2B- 4C- 6D- 8

The correct option is C 6Using 1/λ=R [ 1/n_1^2 1/n_2^2 ]⇒1/975 × 10^ 10 = 1.097 × 10^7 ( 1/1^2 1/n^2 )⇒ n=4 Now number of spectral lines N=n(n 1)/2=4(4 1)/2=6 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Energy Levels

Hydrogen atom...

Question

Hydrogen atom in its ground state is excited by radiation of wavelength $975 \circ A$ . How many lines will be there in
the emission spectrum

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 6
Using $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow \frac{1}{975 \times 10^{- 10}} = 1.097 \times 10^{7} (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow n = 4$
Now number of spectral lines $N = \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

Suggest Corrections

Similar questions

Q. Hydrogen atom in its ground state is excited by radiation of wavelength

975 \circ A

. How many lines will be there in
the emission spectrum

Q. A hydrogen atom in the ground state is excited by radiations of wavelength

975 A

^{o}

. How many lines will be possible in emission spectrum?

Q. A hydrogen atom in the ground state is excited by radiations of wavelength

975 A

^{0}

. The shortest wavelength will be

Q. A hydrogen atom in the ground state is excited by radiations of wavelength 975 A

^{o}

. how many lines will be possible in absorption spectrum

Q. How many hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength

970.6 ˚ A

. How many different lines are possible in the resulting emission spectrum? Find the longest wavelength among these.

Join BYJU'S Learning Program

Hydrogen atom in its ground state is excited by radiation of wavelength 975 ∘A. How many lines will be there in the emission spectrum

The correct option is C 6Using 1λ=R[1n21−1n22]⇒1975×10−10=1.097×107(112−1n2)⇒n=4 Now number of spectral lines N=n(n−1)2=4(4−1)2=6

Hydrogen atom in its ground state is excited by radiation of wavelength $975 \circ A$ . How many lines will be there in
the emission spectrum

The correct option is C 6
Using $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow \frac{1}{975 \times 10^{- 10}} = 1.097 \times 10^{7} (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow n = 4$
Now number of spectral lines $N = \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$