A hydrogen atom is in state N = 4, where N = 1 is the lowest energy state. What is K+U in electron volts for this atomic hydrogen energy state? E4 = eV The hydrogen atom makes a transition to state N = 2. What is K+U in electron volts for this lower atomic hydrogen energy state? E2 = eV What is the energy in electron volts of the photon emitted in the transition from level N = 4 to N = 2? Ephoton = eV

The sum of Kinetic Energy (K) and Potential Energy (U) of the Helium atom is equal to the total energy of Helium atom in the specified state N. From Bohr's atomic model, the energy of a hydrogen atom in state N is given as:

En = K + U = (-1/n²)(13.6 eV)

a)

Here,

n = 4

Therefore,

E₄ = (-1/4²)(13.6 eV)

E₄ = - 0.85 eV

b)

Here,

n = 2

Therefore,

E₂ = (-1/2²)(13.6 eV)

E₂ = - 3.4 eV

c)

The energy of photon emitted in the transition from level 4 to level 2 will be equal to the difference in the energy of both levels:

Ephoton = ΔE = E₄ - E₂

Ephoton = - 0.85 eV - (- 3.4 eV)

Ephoton = 2.55 eV

Cricetus Cricetus · Answer 2 · 2021-11-22T12:46:05+01:00

The electrons energy will be:

[tex]E_4 = -0.85 \ eV[/tex]
[tex]E_2 = -3.4 \ eV[/tex]
[tex]E_{photon} = 2.55 \ eV[/tex]

As we know. the formula:

→ [tex]E_n = K+U[/tex]

[tex]= (-\frac{1}{n^2} ) (13.6 \ eV)[/tex]

(a)

Given:

n = 4

then,

→ [tex]E_4 = (-\frac{1}{4^2} )(13.6 \ eV)[/tex]

[tex]= -0.85 \ eV[/tex]

(b)

Given:

n = 2

then,

→ [tex]E_2 = (-\frac{1}{2^2} )(13.6 \ eV)[/tex]

[tex]= -3.4 \ eV[/tex]

(c)

The energy of photon emitted will be:

→ [tex]E_{photon} = \Delta E[/tex]

[tex]= E_4 -E_2[/tex]

[tex]=-0.85 \ eV-(-3.4 \ eV)[/tex]

[tex]= 2.55 \ eV[/tex]

Thus the above answers are appropriate.

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