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Novay_Z [31]
3 years ago
5

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 ms, on a

verage, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon
Chemistry
1 answer:
elena-s [515]3 years ago
4 0

Answer:

\lambda=3451*10^{10}m

Explanation:

From the question we are told that:

Energy state e=3.50 eV

Time t=2ms

Generally the equation for energy of Photon is mathematically given by

E=e-e_0

E=3.6*10^{-19}J

E=5.7*10^{-19}J

Generally the equation for Wave-length of Photon is mathematically given by

\lambda=\frac{hc}{E}

\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}

\lambda=3451*10^{10}m

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