Question

The wavelength of the electron in the ground state of hydrogen atom is \(y \, \text{\AA}\). What is the wavelength of the electron in the fourth orbit of \(He^+\) ion (in )?

• A. \(2y\)
• B. \(3y\)
• C. \(y\)
• D. \(\dfrac{3y}{2}\)

Hint:

For hydrogen-like species, \[ \lambda \propto \frac{n}{Z} \] where \(n\) is orbit number and \(Z\) is atomic number.

Answer 1

The Correct Option is C

Step 1: Use de-Broglie wavelength relation in Bohr orbit.
For an electron moving in the \(n^{th}\) orbit of a hydrogen-like species, \[ 2\pi r=n\lambda \] From Bohr theory, \[ r_n=\frac{n^2a_0}{Z} \] Hence, \[ \lambda=\frac{2\pi r_n}{n} \] Substituting \(r_n\), \[ \lambda=\frac{2\pi}{n}\left(\frac{n^2a_0}{Z}\right) \] Therefore, \[ \lambda \propto \frac{n}{Z} \]

Step 2: Wavelength in ground state of hydrogen atom.
For hydrogen atom, \[ Z=1 \] Ground state means \[ n=1 \] Thus, \[ \lambda_H \propto \frac{1}{1}=1 \] Given, \[ \lambda_H=y \]

Step 3: Wavelength in fourth orbit of \(He^+\).
For \(He^+\), \[ Z=2 \] Fourth orbit means \[ n=4 \] Therefore, \[ \lambda_{He^+}\propto \frac{4}{2}=2 \] Comparing with hydrogen ground state, \[ \lambda_{He^+}=2y \] But in Bohr’s quantization, \[ n\lambda=2\pi r \] and for hydrogen-like atoms, \[ r_n=\frac{n^2a_0}{Z} \] Thus, \[ \lambda=\frac{2\pi n a_0}{Z} \] For hydrogen ground state: \[ \lambda_H=2\pi a_0 \] For \(He^+\) fourth orbit: \[ \lambda_{He^+}=\frac{2\pi (4)a_0}{2} =4\pi a_0 \] Hence, \[ \lambda_{He^+}=2\lambda_H \] Therefore, \[ \lambda_{He^+}=2y \]

Step 4: Final conclusion.
Hence, the wavelength of electron in the fourth orbit of \(He^+\) ion is \[ \boxed{2y} \]