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Fundamentals Of Physics
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Short Answer

(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?

(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.

See the step by step solution

Step by Step Solution

Step 1: Identification of the given data

The given data is listed below as-

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

Step 2: The energy equation is given by

The energy difference is given by the equation-

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

Here, n is the quantum number.

Step 3: 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=-13.6 eV1n22-1n12

For, n2=3 and n1=2

E=-13.6 eV132-122 =1.889 eV

Now, hc = 1240 eV.nm

λ=hcE =1240 eV.nm1.889 eV =658 nm

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

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

The difference between energies is given by the equation:

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

For, n2= and n1=2

E=-13.6 eV12-122 =3.40 eV

Now, hc = 1240 eV.nm

λ=hcE =1240 eV.nm3.40 eV =366 nm

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

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