Question

The Balmer Rydberg Equation can be extended to ions with only one electron, such
case it has the form
1/lambda=Z^2R(1/m^2-1/n^2) here Z the atomic number. What is the energy of the photon required to
promote an electron in He+ from a 1s orbital to a 2p-orbital
A) 12 hoR
B) 6 hoR
C)3 HCR
D)3/4hcR
please answer and no spamming please I will mark as brainliest ​

Answer 1

Answer:

C

Explanation:

Answer 2

Answer:

The energy of the emitted photon which is required to promote an electron from a 1s-orbital to a 2p-orbital in He⁺ is 3hcR.

Therefore, option c) 3hcR is the correct option.

Explanation:

Given,

An electron in He⁺ is promoted from a 1s-orbital to a 2p-orbital.

Therefore,

The radial quantum number of the initial state, m = 1

The radial quantum number of the final state, n = 2

The energy of the photon required to promote an electron in He⁺ from a 1s-orbital to a 2p-orbital, E =?

As we know,

• The Balmer-Rydberg equation can be extended to ions with only one electron, like He⁺.
• The Rydberg equation is used to calculate the wavelength of element spectral lines as given below:
• $\frac{1}{\lambda} = RZ^{2} [\frac{1}{m^{2} } - \frac{1}{n^{2} } ]$

Here,

• λ = The wavelength of the emitted photon
• R = The Rydberg constant
• Z = The atomic number of the atom( He⁺ ) = 2
• m = The radial quantum number of the initial state
• n = The radial quantum number of the final state

After putting all the values in the Balmer-Rydberg equation, we get:

• $\frac{1}{\lambda} = R(2)^{2} [\frac{1}{(1)^{2} } - \frac{1}{(2)^{2} } ]$
• $\frac{1}{\lambda} = 4R [\frac{1}{1 } - \frac{1}{4 } ]$
• $\frac{1}{\lambda} = 4R [\frac{4-1}{4} ]$
• $\frac{1}{\lambda} = 3R$

Hence, λ = $\frac{1}{3R}$

Now, as we know that the frequency (v) can be calculated by the equation:

• $v = \frac{c}{\lambda}$

Here,

• v = The frequency of the emitted photon
• c = Speed of the light
• λ = The wavelength of the emitted photon

After putting the value of λ in the equation, we get:

• $v = \frac{c}{1/3R}$
• $v = 3Rc$

As we know,

• The energy of the photon is calculated by the equation:
• E = hv

Here,

• E = The energy of the photon
• h = The Planck's constant
• v = The frequency of the emitted photon

After putting the value of v in the equation, we get:

• E = h3Rc = 3hcR

Therefore, the energy of the photon is 3hcR.