Question

The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is

• A. \(0.3 A˚\)
• B. \(3.3  A˚\)
• C. \(6.26  A˚\)
• D. \(10 A˚\)

Answer 1

The Correct Option is B

De-Broglie wavelength is given by
$\lambda=12.27 / \sqrt{ E }_{0}$,
where $E _{0}$ is the ground state energy of the hydrogen atom
whose value is $13.6\, V$
$\therefore \lambda=12.27 /(\sqrt{1} 3.6)=3.33 \,?$

Answer 2

The de-Broglie wavelength \( \lambda \) associated with an electron is given by the formula: \[ \lambda = \frac{h}{mv} \] Where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J·s} \)), - \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)), - \( v \) is the velocity of the electron. For a hydrogen atom in its ground state, the velocity of the electron is given by the Bohr model of the atom: \[ v = \frac{2 \pi k e^2}{h} \] Using the known values for constants and the radius of the electron's orbit in the ground state (\( r = 5.3 \times 10^{-11} \, \text{m} \)), we can calculate the wavelength of the electron. The result gives: \[ \lambda = 3.3 \, \text{Å} \] Thus, the de-Broglie wavelength associated with the electron in the ground state of the hydrogen atom is 3.3 Å.