CBSE Class 12 · 2016 · All India · Modern PhysicsShort Answer

When is the

H_{α}

line in the emission spectrum of a hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.

Answer: Balmer line, $n = 3 \to 2$ ; $ν \approx 4.57 \times 1 0^{14} Hz$ .

Solution

The

H_{α}

line is the first line of the Balmer series, emitted when the electron jumps from

n = 3

n = 2

Δ E = 13.6 (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = 13.6 \times \frac{5}{36} = 1.89 eV = 3.03 \times 1 0^{- 19} J .

ν = \frac{Δ E}{h} = \frac{3.03 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} \approx 4.57 \times 1 0^{14} Hz .

ν \approx 4.57 \times 1 0^{14} Hz (n = 3 \to 2)

CBSE Class 12 Board · 2016 · All India · Subjective