Calculate the energy of the photon emitted for transition a


Asked by maham237 @ in Chemistry viewed by 334 People


Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=7 to n=1

Answered by maham237 @



Answer : The energy of photon = 21.369\times 10^{-19}J

Solution : Given,

Final energy level = 1

Initial energy level = 7

Formula used :

\frac{1}{\lambda}=R[ \frac{1}{n^2_{final}}- \frac{1}{n^2_{initial}}] ..........(1)

where,

\lambda = wavelength

R = Rydberg's constant = 1.0974 \times 10^7m^{-1}

n_{final} = final energy level

n_{initial} = initial energy level

Now put all the given values in above formula (1), we get the value of wavelength.

\frac{1}{\lambda}=1.0974\times 10^7m^{-1} [\frac{1}{1^2}- \frac{1}{7^2}] = 1.0750\times 10^7m^{-1}

\lambda=0.9302\times 10^{-7}m

Now we have to calculate the energy of photon by using formula,

E=\frac{h\times c}{\lambda}

where,

E = energy

h = planck's constant = 6.626\times 10^{-7}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength

put all the values in this formula, we get the energy of photon.

E=\frac{6.626\times 10^{-7}Js\times 3\times 10^8m/s}{0.9302\times 10^{-7}m} = 21.369\times 10^{-19}J



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