A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav

use the eqn:
f=R*{(1/n1)^2~(1/n2)^2},
R=Rydberg constant=109678/cm
n=energy level number
and then c=f? for the wavelength.

the electron in 5th energy level have energy=-0.54 eV
the electron in 2nd energy level have energy= -3.4eV
thus energy released when it jumps from 5th to 2nd state=
     -0.54-(-3.4) = 2.86eV
 and 1eV = 1.6 x 10^-19 joule
       2.86eV= 4.58 x 10^-19 joule
  from planck's formula
  hv = energy radiated = 4.58 x 10^-19
 v=  6.19 x 10^14 hertz
and wavelength = 4.18 x 10^-7 meter

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First, find the energy of each level:

R=1.0974 x 10^7 m^-1
h= Planck's constant= 6.626 x 10^-34 J*s
c=speed of light=3.0 x 10^8 m/s


Energy level=[ -(R)(h)(c)] / (level # squared)

Energy level 2= [ -(1.0974 x 10^7)(6.626x 10^-34)(3.0 x 10^8)] / 4
= -5.45 x 10^-19 J
 
Energy level 5 = [ -(1.0974 x 10^7)(6.626x 10^-34)(3.0 x 10^8)] / 25
= -8.72 x 10^-20 J

Then take the absolute value of the difference of the two levels' energies.

|(-5.45 x 10^-19 J) - (-8.72 x 10^-20 J)|= 4.578 x 10^-19 J

Then use that to find the frequency using the eqtn
Energy=(h)(frequency)

So
4.578 x 10^-19 = (6.626 x 10^-34)(frequency)
frequency= 6.909 x 10^14

And then use the frequency to solve for the wavelength where
(wavelength)(frequency)=speed of light

So then wavelegth= (3.0 x 10^8) / (6.909 x 10^14)
Which gives us  4.34 x 10^-7 m.
Then convert to nm, which gives you 434 nm as your wavelength.

several people have shown you the steps to find the frequency and wavelength of the photon, one note that might be helpful depending on your course is that for the hydrogen atom, transitions into the second energy level have a special name, these are called Balmer lines and are the visible lines in the spectrum of hydrogen

transitions into the first energy level are called the Lyman lines and are of higher energy and are in the ultraviolet part of the spectrum

transitions into the third and higher levels have less energy and are in the infrared, so the Balmer series are the only transitions that produce visible light