SATHEE: Atomic Structure Question 194

Atomic Structure Question 194

Question: The energy of electron in first Bohr’s orbit of H-atom is -13.6 eV. What will be its potential energy in n = $4^{t h}$ orbit

Options:

A) -13.6eV

B) -3.4eV

C) -0.85eV

D) -1.70eV

Show Answer

Answer:

Correct Answer: D

Solution:

Energy of nth orbit of H-atom $= - \frac{2 π^{2} m e^{4} k^{2}}{h^{2}} \times \frac{1}{h^{2}}$ Energy of Boht’s orbit of H-atom $= - \frac{2 π^{2} m e^{4} k^{2}}{h^{2}}$

$= - 13.6 e V$ (given) Energy of fourth Bohr’s of H-atom $= - \frac{2 π^{2} m e^{4} k^{2}}{h^{2}} \times \frac{1}{4^{2}}$

$= 13.6 \times \frac{1}{16} e V = - 0.85 e V$ PE of electron in nth orbit = $2 \times E_{n}$ So P.E. of electron in $4^{t h}$ orbit $= 2 \times (- 0.85) = - 1.70 e V$

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Atomic Structure Question 194

Question: The energy of electron in first Bohr’s orbit of H-atom is -13.6 eV. What will be its potential energy in n = 4th orbit

Options:

Answer:

Solution:

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Question: The energy of electron in first Bohr’s orbit of H-atom is -13.6 eV. What will be its potential energy in n = $4^{t h}$ orbit