An electron in hydrogen atom in second excited state jumps to the first and ground state of the atom. Find the ratio of wavelengths emitted during this process
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Answer:
The ratio of wavelengths emitted during this process will be 5/27
Explanation:
The energy equation of the wave emitted is given by,
hc/λ = -13.6[ 1/n₂² - 1/n₁²]
when the electron jumps from second state to the first state,
n₂ = 3 and n₁ = 2
Hence ,
hc/λ₁ = -13.6[1/3² - 1/2²]
=> hc/λ₁ = -13.6[ 1/9 - 1/4]
=> hc/λ₁ = -13.6[-5/36]............eqn1
Similarly, when the electron jumps from first state to the ground state,
n₂ = 2 and n₁ = 1
Hence ,
hc/λ₂ = -13.6[1/2² - 1/1²]
=> hc/λ₂ = -13.6[ 1/4 - 1]
=> hc/λ₂ = -13.6[-3/4]...................eqn2
dividing eqn 1 by eqn 2 we get
λ₂/λ₁ = (5/36)/(3/4) = 5/27
Hence the ratio of wavelengths emitted during this process will be 5/27
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