What is the wavelength, in nm, of a photon emitted during a transition from the n = 5 state to the n = 2 state in the hydrogen atom ?

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71
Jul 31, 2017

This should be a transition in the so called "Balmer Series":

enter image source here
(Picture from Ohanian Physics)

You can use the fact that a photon emitted during the transition from n = 5 to n = 2 will carry an energy #E# equal to the difference between the energies of these two states.

Knowing this, you can relate the energy of the photon with the frequency #nu# through

#E=hnu#

(#h# is Planck's constant.)

For any state corresponding to #n# in the hydrogen atom, you get

#E_n= -"13.6 eV"/n^2#,

where #-"13.6 eV"# is the approximate ground-state energy of the hydrogen atom.

So:

#E_2 = -"13.6 eV"/4 = -"3.4 eV" = -5.44*10^-19# #"J"#

#E_5 = -"13.6 eV"/25 = "0.544 eV" = -8.7*10^-20# #"J"#

So,

#DeltaE = 4.57*10^-19# #"J"#.

In #E=hnu#,

#nu=(4.57*10^-19 "J")/(6.63*10^-34 "J" cdot"s") = 6.892*10^14 "s"^(-1)#,

but #c=lambdanu# (#c# is the speed of light);

So,

#lambda=(3*10^8 "m/s")/(6.892*10^14 "s"^(-1)) xx (10^9 "nm")/("1 m") = ul"435 nm"#

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Oct 23, 2017

Answer:

#lambda=433.936 7nm#

Explanation:

#lambda=(hc)/E_p#

#E_p# is the energy of photon.

Since, #E_p=(2pi^2m_eK^2Z^2e^4)/h^2(1/n_2^2-1/n_5^2)#

#K=1/(4piepsilon_0), n_2=2, n_5=5, Z=1#

#lambda=(800*h^3*c*epsilon_0^2)/(21*m_e*e^4)#

Putting in the values of physical constants

#lambda=433.9367 nm#

Here, #lambda=#wavelength of photon

#h=#Planck's constant

#m_e=#mass of electron

#c=#speed of light in vacuum

#e=#elementary charge

#Z=#Atomic No. [1 for hydrogen]

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