Question

Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with Z=5 and only a single electron orbiting the nucleus. This expression predicts equal electron energies for these two species for certain values of the quantum number n (the quantum number is different for each species). For quantum numbers less than or equal to 9, what is the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level?

Answer 1

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

so

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= -  13.6 x2² / 8²

= -  0.85 eV .