# If light of frequency #6.90 xx 10^14 "s"^(-1)# is emitted when an electron in hydrogen atom relaxes from #n_1 = 5#, what is #n_2#?

##### 1 Answer

#### Explanation:

All you have to do here is use the **Rydberg equation** for hydrogen, which looks like this

#color(blue)(ul(color(black)(1/(lamda) = R * (1/n_2^2 - 1/n_1^2))))#

Here

#lamda# is thewavelengthof the emitted photon#R# is the Rydberg constant, equal to#~~1.097 * 10^(7)"m"^(-1)# #n_1# is the principal quantum number that describes the orbitalfrom whichthe transition is made#n_2# is the principal quantum number that describes the orbitalto whichthe transition is made

Now, notice that the problem provides you with the **frequency** of the emitted photon, so make sure that you convert it to wavelength by using the equation

#color(blue)(ul(color(black)(lamda * nu = c)))#

Here

#nu# is thefrequencyof the photon#c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)#

In your case, you will have

#lamda = c/(nu)#

#lamda = (3 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))))/(6.90 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))) = 4.348 * 10^(-7)"m"#

Rearrange the Rydberg equation to solve for

#1/(lamda) = R * (1/n_2^2 - 1/n_1^2)#

#1/(lamda) * 1/R = (n_1^2 - n_2^2)/(n_1^2 * n_2^2)#

This is equivalent to

#n_1^2 * n_2^2 = lamda * R * n_1^2 - lamda * R * n_2^2#

and so

#n_1^2 * n_2^2 + lamda * R * n_2^2 = lamda * R * n_1^2#

#n_2^2 * (n_1^2 + lamda * R) = lamda * R * n_1^2#

#n_2^2 = (lamda * R * n_1^2)/(n_1^2 + lamda * R)#

Finally,

#n_2 = sqrt( (lamda * R * n_1^2)/(n_1^2 + lamda * R)#

You know that

#n_2 = sqrt( (4.348 * color(blue)(cancel(color(black)(10^(-7)))) color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^(7))))color(red)(cancel(color(black)("m"^(-1)))) * 5^2)/(5^2 + 4.348 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^(7))))color(red)(cancel(color(black)("m"^(-1))))))#

#color(darkgreen)(ul(color(black)(n_2 = sqrt(4.00554) ~~ 2)))#

Therefore, you can say that the frequency of the emitted photon corresponds to an **Balmer series**.

As you can see, the spectral line associated with this transition is located in the *visible region* of the **electromagnetic spectrum**, the emitted photon corresponding to the color **blue**.