Question

If the ground state energy of the electron in a hydrogen atom is -13.6 ev, then what would be the energy of the electron in 2nd bohr orbit of a he + ion?

Answer 1

Answer:the energy in the second orbit of He atom will be : -13.6 ev

Explanation:

Answer 2

Answer:

Complete step by step solution:

This numerical is based on the energy of the electron given by Bohr. According to the Bohr, the electrons revolve around the nucleus in that fixed orbit and has a definite value of energy and these orbits are also known as energy levels and the energy of the electrons in that particular orbit is fixed and doesn’t change with time. The different energy levels are numbered as 1,2,3,4,5---etc. starting from the nucleus. The energy of each orbit is given by the expression as:

En=−2π2me4n2h2

Substituting the values, the values of mSubstituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get:

En=−21.8×10−19n2J/atom

⇒ −1312n2kJmole−1

⇒ −13.6n2eV/atom (1eV=1.602 ×J)

Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.

But for hydrogen like particles, the expression for energy is:

En=−2π2mZ2e4n2h2⇒ −1312Z2n2kJmole−1⇒ −2.18×10−11n2ergsNow considering the statement;

We can find the energy of the hydrogen electron by applying the formula as;

En=−2.18×10−11n2ergs

Energy of the hydrogen atom in first Bohr orbit E1=−2.18×10−1112=−2.18×10−111ergs

Energy of the hydrogen atom in fifth Bohr orbit E5=−2.18×10−1152=−2.18×10−1125ergs

Energy required to shift the electron of the hydrogen atom from first to fifth Bohr orbit is as;

ΔE=E5−E1⇒ −2.18×10−1125−2.18×10−111⇒ 2.18×10−11(11−125)⇒ 2.18×10−11(25−125)⇒ 2.18×10−11(2425)⇒ 2.09×10−11 ergs⇒ 2.09×10−18J (1 erg=107J)So, the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit is 2.09×10−18J.

Now, calculating the wavelength of the light emitted as;

As we know;

E=hv⇒ hcλ (λ=cv) λ=hcE --------------(1)

So, as we know that;

E=2.09×10−11 ergs

h=6.62×10−27 erg sec−1

c=3×1010 cm sec−1

then, put all these values in equation (1), we get;

λ=6.62×10−27 erg sec−1 × 3×1010 cm sec−12.09×10−11 ergs⇒ 9.51×10−6 cm

Hence, the wavelength of the light emitted when the electron returns to the ground state is 9.51×10−6 cm.