Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
A detailed answer please...
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Answered by
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According to Balmer formula
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
hope it helps D:
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
hope it helps D:
spidermangamer426:
please mark it as brainilst
Which value have you taken for RH ?
see in the answer
You have not mentioned
What is RH's value here? Is it Rydberg's const. or what???
yes it is
its value is 109677 cm^-1
Okay...Thank You !
Answered by
3
Answer:
According to Balmer formula
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
hope it helps D:
Explanation:
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