Question

calculate the
wavenumber for the longest wave Length
transition in Balmer series of hydrogen. R = 1.09x 107m-1
[hint : longest wavelength radiation corresponds to
smallest frequency
and least energy
Take n1= 2,
n2=3, R (Rh) = 1.09 x 107 m-1​

Answer 1

Answer:

Explanation:

Thanks for free points

Answer 2

Explanation:

According to Balmer formula

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]For the Balmer series, ni = 2.

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]For the Balmer series, ni = 2.Thus, the expression of wavenumber(ṽ) is given by,

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]For the Balmer series, ni = 2.Thus, the expression of wavenumber(ṽ) is given by,Wave number (ṽ) is inversely proportional to the wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]For the Balmer series, ni = 2.Thus, the expression of wavenumber(ṽ) is given by,Wave number (ṽ) is inversely proportional to the 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:

According to Balmer formulav = λ1 =RH [ n 121 − n 221 ]For the Balmer series, ni = 2.Thus, the expression of wavenumber(ṽ) is given by,Wave number (ṽ) is inversely proportional to the 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