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defon
2 years ago
7

Calculate the energy for the transition of an electron from the n = 2 level to the n = 5 level of a hydrogen atom.

Chemistry
1 answer:
katovenus [111]2 years ago
4 0

The energy for the transition:

The energy for the transfer of an electron from the n=2 level to the n=5 level of a hydrogen atom is 4.57 x 10^{-19} J

Calculation:

We are aware of the energy levels as n=2 and n=5. So, we may determine the wavelength of the photon released by the electron during this transition using Rydberg's equation:

1/λ = R x (1/n^{2}_{final} - 1/n^{2} _{initial})

where,

1/λ = the wavelength of the emitted photon,

R = Rydberg's constant, 1.0974 x 10^{7} m^{-1}

n_{final} = the final energy level = 5

n_{initial} = the initial energy level = 2

Now, put the value in the above equation, we get,

1/λ = 1.0974 x 10^{7} m^{-1} x ( 1/5^{2} - 1/2^{2} )

1/λ = 1.0974 x 10^{7}m^{-1} x (-0.21)

1/λ = -0.23 x 10^{7}m^{-1}

λ = 4.347 x 10^{-7}m

Since, E = hc/λ

where,

h = Plank's constant = 6.626 x 10^{-34} Js

c = speed of light = 3 x 10^{8} m/s

So, the transition energy for your particular transition is,

E = 6.626 x 10^{-34} x 3 x 10^{8} / (4.347 x 10^{-7})

E = 4.57 x 10^{-19} J

Learn more about Rydberg's formula here,

brainly.com/question/13185515

#SPJ4

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In the rate of reaction, A and B are the reactants and C and D are the products.

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The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

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\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

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