# 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":

(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

$E = h \nu$

($h$ is Planck's constant.)

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

${E}_{n} = - \frac{\text{13.6 eV}}{n} ^ 2$,

where $- \text{13.6 eV}$ is the approximate ground-state energy of the hydrogen atom.

So:

${E}_{2} = - \text{13.6 eV"/4 = -"3.4 eV} = - 5.44 \cdot {10}^{-} 19$ $\text{J}$

${E}_{5} = - \text{13.6 eV"/25 = "0.544 eV} = - 8.7 \cdot {10}^{-} 20$ $\text{J}$

So,

$\Delta E = 4.57 \cdot {10}^{-} 19$ $\text{J}$.

In $E = h \nu$,

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

but $c = \lambda \nu$ ($c$ 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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8
Oct 23, 2017

$\lambda = 433.936 7 n m$

#### Explanation:

$\lambda = \frac{h c}{E} _ p$

${E}_{p}$ is the energy of photon.

Since, ${E}_{p} = \frac{2 {\pi}^{2} {m}_{e} {K}^{2} {Z}^{2} {e}^{4}}{h} ^ 2 \left(\frac{1}{n} _ {2}^{2} - \frac{1}{n} _ {5}^{2}\right)$

$K = \frac{1}{4 \pi {\epsilon}_{0}} , {n}_{2} = 2 , {n}_{5} = 5 , Z = 1$

$\lambda = \frac{800 \cdot {h}^{3} \cdot c \cdot {\epsilon}_{0}^{2}}{21 \cdot {m}_{e} \cdot {e}^{4}}$

Putting in the values of physical constants

$\lambda = 433.9367 n m$

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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