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?

Answers: 3

Show answers

Answers

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:

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

So, our line is contained in the n=3 series.

E) Ultraviolet

We can answer this question by looking at the different wavelengths of the electromagnetic spectrum. In fact, we have:

Ultraviolet: 380 nm - 1 nm

Visible: 750 nm - 380 nm

Infrared: 1 mm - 750 nm

Our wavelength here is

97.26 nm

So, we see it is included in the ultraviolet part of the spectrum. In fact, all lines in the Lyman series (n=1) lie in the ultraviolet ragion.

F) Infrared

Again, the electromagnetic spectrum is:

Ultraviolet: 380 nm - 1 nm

Visible: 750 nm - 380 nm

Infrared: 1 mm - 750 nm

Our wavelength here is

1282 nm

So, we see it is included in the infrared part of the spectrum. In fact, all lines in the Paschen series (n=3) lie in the infrared band.

Answer from: Quest

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

Answer from: Quest

Football was the fourth deadliest sport with a fatality rate of .81 per 100,000 participants. cheerleading, which has been shown to be the most dangerous activity based on injury data, was not included in the study because it was not classified as a sport.this is my answer because it didn’t give me any sports

Another question on Physics

Physics, 21.06.2019 17:30

Abatter hits the baseball a with an initial velocity of v0 = 110 ft/sec directly toward fielder b at an angle of 23° to the horizontal; the initial position of the ball is 2.2 ft above ground level. fielder b requires 0.44 sec to judge where the ball should be caught and begins moving to that position with constant speed. because of great experience, fielder b chooses his running speed so that he arrives at the “catch position” simultaneously with the baseball. the catch position is the field location at which the ball altitude is 8.4 ft. determine the velocity of the ball relative to the fielder at the instant the catch is made.

Answers: 1

Answer

Physics, 22.06.2019 05:30

Física: un futbolista patea hacia el arco con una velocidad de 15m/ah, calcular: a)el alcance para un ángulo de tiro de 25? , b) el tiempo que el balón permanece en el aire.

Answers: 3

Answer

Physics, 22.06.2019 07:00

If a tank filled with water contains a block and the height of the water above point a within the block is 0.8 meter, what is the pressure at point a?

Answers: 2

Answer

Physics, 22.06.2019 11:00

What would be the result of an alpha particle coming into a magnetic field? a) the alpha particle will stop moving. b) the alpha particle will reverse its direction. c) the alpha particle will be deflected in a curve path. d) the alpha particle will continue to travel in a straight line.

Answers: 1

Answer

You know the right answer?

Two of the wavelengths emitted by a hydrogen atom are 97.26 nm and 1282 nm . part a what is the m va...

Questions