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(a) What is the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines? (b) What is the wavelength of the series limit?

Short Answer

Expert verified

(a)The wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines is 658 nm.

(b) Thus, the wavelength of the series limit is 366 nm.

Step by step solution

01

Identification of the given data

The given data is listed below as-

The photon emitted in the Balmer series is the least energetic.

02

The energy equation is given by

Theenergy differenceis given by the equation-

E=(13.6eV)(1n22-1n12)

Here, n is the quantum number.

03

To determine the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines (a)

The difference between energies is given by the equation:

E=E3-E2E=-13.6eV1n22-1n12

For, n2=3and n1=2

E=-13.6eV132-122=1.889eV

Now, hc = 1240 eV.nm

λ=hcE=1240eV.nm1.889eV=658nm

Thus, the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines is 658 nm.

04

Step 4: To determine the wavelength of the series limit. (b)

The difference between energies is given by the equation:

E=E-E2E=-13.6eV1n22-1n12

For, n2=and n1=2

E=-13.6eV12-122=3.40eV

Now, hc = 1240 eV.nm

λ=hcE=1240eV.nm3.40eV=366nm

Thus, the wavelength of the series limit is 366 nm.

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