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In his experiment, Balmer introduced hydrogen gas in a discharged tube in order to excite electrons on higher energy levels. Then, as the electrons dropped down to lower energy levels, photons were emitted by the atoms. What is the energy of a photon emitted by a electron dropping from level n

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Answer:

2.56 eV

Explanation:

Using the formula;

ΔE= -RH(1/n^2final - 1/n^2initial)

RH = -2.18 * 10-18 J

nfinal = 2

ninitial =1

Substituting values;

ΔE= -2.18 * 10-18 J(1/4^2 - 1/2^2)

ΔE= -2.18 * 10-18 J(0.0625 - 0.25)

ΔE= 4.09 * 10^-19 J

To convert to eV

4.09 * 10^-19 J/1.6 * 10^-19

=2.56 eV

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