Calculate the wavelength in nanometers of the light emitted by a hydrogen atom w

Question

Calculate the wavelength in nanometers of the light emitted by a hydrogen atom when it's electron falls from the n= 7 to the n= 4 principal energy level. Recall that the engergy levels of the H atom are given by En = -2.18 x 10 to the negative 18 J ( 1/n to the second power) . ( c= 3.00 x 10 to to the 8 th power m/s ; h=6.63 x 10 to the negative 34 j.s Calculate the wavelength in nanometers of the light emitted by a hydrogen atom when it's electron falls from the n= 7 to the n= 4 principal energy level. Recall that the engergy levels of the H atom are given by En = -2.18 x 10 to the negative 18 J ( 1/n to the second power) . ( c= 3.00 x 10 to to the 8 th power m/s ; h=6.63 x 10 to the negative 34 j.s

Explanation / Answer

Given : n = 7 , n = 4

We use energy equation

We know

Delta E = –2.18 E -18 J (1/nf2– 1/ni2)

We know ni is 7 and nf is 4

Lets plug these value in energy equation

Delta E = –2.18 E -18 J (1/42– 1/72)

Delta E = –9.18 E-20 J

Calculation of wavelength:

Delta E = hc/

Here h is planks constant = 6.63 E -34 J.s , c is speed of light = 3.00 E 8 m/s

. is wavelength

Lets rearrange the equation ;

. = hc / Delta E

Lets plug all the value in above equation

. = (6.63 E -34 J.s * 3.0 E 8 m/s ) / 9.18 E -20 J

= 2.17 E -6 m

Conversion of wavelength into nm

= 2.17 E -6 m * 1 E9 nm / 1 m

= 2167.6 nm

. = 2167.6 nm

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