Physics, 29.01.2020 19:46 sunshine52577oyeor9

Two of the wavelengths emitted by a hydrogen atom are 97.26 nm and 1282 nm . part a what is the m value for the wavelength 97.26 nm ? express your answer as an integer. part b what is the n value for the wavelength 97.26 nm ? express your answer as an integer. part c what is the m value for the wavelength 1282 nm ? express your answer as an integer. part d what is the n value for the wavelength 1282 nm ? express your answer as an integer. part e part complete for the wavelength 97.26 nm , is the light infrared, visible, or ultraviolet? part f part complete for the wavelength 1282 nm , is the light infrared, visible, or ultraviolet?

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Answer from: Danielyanez

A) m = 4

We can solve the problem by using Rydberg equation:

$\frac{1}{\lambda}=R_H (\frac{1}{n^2}-\frac{1}{m^2})$

where

$R_H = 1.097\cdot 10^7 m^{-1}$ is the Rydberg constant for hydrogen

n is the principal quantum number of the upper energy level

m is the principal quantum number of the lower energy level

For the first wavelength, we have

$\lambda=97.26 nm = 97.26\cdot 10^{-9} m$

Substituting into the equation, we find

$\frac{1}{n^2}-\frac{1}{m^2}=\frac{1}{\lambda R_H}=\frac{1}{(97.26\cdot 10^{-9} m)(1.097\cdot 10^7 m^{-1})}=0.9373$ )

By setting n=1, we obtain the Lyman series which goes from 121.6 nm (for m=2) to 91.18 nm (for $m=\infty$ ). So our line of 97.26 nm must be in this series.

By setting n=1, we find m:

$\frac{1}{m^2}=\frac{1}{n^2}-0.9373=\frac{1}{1^2}-0.9373=0.0627\\m=\frac{1}{\sqrt{0.0627}}=4$

B) n = 1

n can be found by thinking about the limit of the different series.

Larger n corresponds to larger wavelengths; for each n, m goes from (n+1) to $\infty$ , and the shortest wavelength of each series is the one corresponding to $m=\infty$ .

If we put n = 2, and $m=\infty$ , we find the shortest wavelength of the n=2 series:

$\frac{1}{\lambda}=R_H (\frac{1}{n^2}-\frac{1}{m^2})=(1.097\cdot 10^7 m^{-1})(\frac{1}{2^2}-\frac{1}{\infty})=\frac{1.097\cdot 10^7 m^{-1}}{4}=2.74\cdot 10^6 m^{-1}\\\lambda=\frac{1}{2.74\cdot 10^6 m^{-1}}=3.64\cdot 10^{-7} m = 364 nm$

which is longer than our line at 97.26 nm, so n must be smaller than 2, which means n=1.

C) m = 5

Similarly to what we did in part A), here we have a wavelength of

$\lambda=1282 nm = 1282\cdot 10^{-9} m$

Substituting into the Rydberg equation, we find

$\frac{1}{n^2}-\frac{1}{m^2}=\frac{1}{\lambda R_H}=\frac{1}{(1282\cdot 10^{-9} m)(1.097\cdot 10^7 m^{-1})}=0.0711$ )

By setting n=3, we obtain the Paschen series which goes from 1875 nm (for m=4) to 820.4 nm (for $m=\infty$ ). So our line of 1282 nm must be in this series.

By setting n=3, we find m:

$\frac{1}{m^2}=\frac{1}{n^2}-0.0711=\frac{1}{3^2}-0.0711=0.04001\\m=\frac{1}{\sqrt{0.04001}}=5$

D) n = 3

Similarly to what we did in part B), if we put n = 4, and $m=\infty$ , we find the shortest wavelength of the n=4 series:

$\frac{1}{\lambda}=R_H (\frac{1}{n^2}-\frac{1}{m^2})=(1.097\cdot 10^7 m^{-1})(\frac{1}{4^2}-\frac{1}{\infty})=\frac{1.097\cdot 10^7 m^{-1}}{16}=6.856\cdot 10^5 m^{-1}\\\lambda=\frac{1}{6.856\cdot 10^5 m^{-1}}=1.458\cdot 10^{-6} m = 1458 nm$

which is longer than our line at 1282 nm, so n must be smaller than 4. Indeed, if we try with n=3, we find:

$\frac{1}{\lambda}=R_H (\frac{1}{n^2}-\frac{1}{m^2})=(1.097\cdot 10^7 m^{-1})(\frac{1}{3^2}-\frac{1}{\infty})=\frac{1.097\cdot 10^7 m^{-1}}{9}=1.219\cdot 10^6 m^{-1}\\\lambda=\frac{1}{1.219\cdot 10^6 m^{-1}}=8.204\cdot 10^{-7} m = 820.4 nm$