Question

In the ground state of hydrogen atom,it's bohr radius is given as 5.3×10 to the power -11 m. The atom is ecited such that the radius become 21.2*10 raise to the power -11 m.find the value of the principal quantum number and the total energy of the atom in this excited state

Answer 1

Answer:

Principal quantum number is n = 2 and total energy is E = -3.4 eV

Explanation:

As we know that radius of nth excited state as per Bohr's model is given as

$r = r_o \frac{n^2}{z}$

where we know that

$r_o = 5.3 \times 10^{-11} m$

now we have

$r = 21.2 \times 10^{-11}m$

so we have

$21.2 \times 10^{-11} = (5.3 \times 10^{-11}) n^2$

$n^2 = 4$

$n = 2$

so principal quantum number is n = 2

Total energy is given as

$E = -13.6 \frac{z^2}{n^2} eV$

$E = -13.6 \frac{1}{4} eV$

$E = -3.4 eV$

#Learn

Topic : Bohr's model

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