Consider a hydrogen atom with its electron in the nth orbital. An elec...
Given data:
Wavelength of electromagnetic radiation, λ = 90 nm
Kinetic energy of ejected electron, KE = 10.4 eV
hc = 1242 eV nm
To find: Value of n in the hydrogen atom
Solution:
We know that the energy of the photon is given by the formula:
E = hc/λ
where,
h = Planck's constant = 6.626 x 10^-34 J s
c = speed of light = 3 x 10^8 m/s
λ = wavelength of the radiation
Let's first convert the wavelength given in nanometers to meters:
λ = 90 nm = 90 x 10^-9 m
Substituting the values in the formula, we get:
E = (6.626 x 10^-34 J s x 3 x 10^8 m/s) / (90 x 10^-9 m)
E = 2.21 x 10^-18 J
We know that the energy required to ionize a hydrogen atom from its nth energy level is given by:
En = -13.6/n^2 eV
where,
En = energy of the nth energy level
n = principal quantum number
To completely ionize the atom, the energy of the photon must be equal to or greater than the energy required to remove the electron from the atom.
So, we can write:
E ≥ En
Substituting the values, we get:
2.21 x 10^-18 J ≥ (-13.6/n^2) eV
Let's convert the kinetic energy of the ejected electron from eV to joules:
1 eV = 1.6 x 10^-19 J
KE = 10.4 eV = 1.6 x 10^-19 J x 10.4 = 1.664 x 10^-18 J
Now, using the above equation, we get:
1.664 x 10^-18 J ≥ (-13.6/n^2) eV
Dividing both sides by -13.6 eV, we get:
-1.224 x 10^-19 ≥ 1/n^2
Taking the reciprocal of both sides and multiplying by -1, we get:
n^2 ≥ 819.18
n ≥ √819.18
n ≥ 28.6
Since the value of n must be an integer, the only possible value of n is 2.
Therefore, the value of n in the hydrogen atom is 2.
Consider a hydrogen atom with its electron in the nth orbital. An elec...