Electron in hydrogen atom is excited in higher energy level where is stay for 10 to power -8 second makes 5.25 ×10 to power 5 revolution.find de Broglie wavelength of the electron in same quantum level of He + ion

To find the de Broglie wavelength of an electron in the same quantum level of a He⁺ ion, we first need to understand a few key concepts about quantum mechanics and the behavior of electrons in atoms.

Understanding de Broglie Wavelength

The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:

\( \lambda = \frac{h}{p} \)

where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) and \( p \) is the momentum of the particle. For an electron, momentum can be expressed as:

\( p = mv \)

where \( m \) is the mass of the electron and \( v \) is its velocity.

Calculating the Velocity of the Electron

In a hydrogen-like atom such as He⁺, the electron's velocity can be derived from its energy level. The energy of an electron in a hydrogen-like atom is given by:

\( E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \)

Here, \( Z \) is the atomic number (for He, \( Z = 2 \)), and \( n \) is the principal quantum number. For the first excited state, \( n = 2 \).

Substituting the values:

\( E_2 = -\frac{2^2 \cdot 13.6}{2^2} = -3.4 \, \text{eV} \)

To convert this energy into joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\( E_2 = -3.4 \times 1.6 \times 10^{-19} = -5.44 \times 10^{-19} \, \text{J} \)

Finding the Velocity

The kinetic energy of the electron can be expressed as:

\( KE = \frac{1}{2} mv^2 \)

Setting the kinetic energy equal to the absolute value of the energy (since we are interested in the magnitude), we have:

\( \frac{1}{2} mv^2 = 5.44 \times 10^{-19} \)

The mass of the electron (\( m \)) is approximately \( 9.11 \times 10^{-31} \, \text{kg} \). Rearranging the equation to solve for \( v \):

\( v = \sqrt{\frac{2 \cdot 5.44 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 1.16 \times 10^7 \, \text{m/s} \)

Calculating the de Broglie Wavelength

Now that we have the velocity, we can calculate the momentum:

\( p = mv = 9.11 \times 10^{-31} \cdot 1.16 \times 10^7 \approx 1.06 \times 10^{-23} \, \text{kg m/s} \)

Substituting this value into the de Broglie wavelength formula:

\( \lambda = \frac{6.626 \times 10^{-34}}{1.06 \times 10^{-23}} \approx 6.25 \times 10^{-11} \, \text{m} \)

Final Result

Thus, the de Broglie wavelength of the electron in the same quantum level of the He⁺ ion is approximately:

\( \lambda \approx 6.25 \times 10^{-11} \, \text{m} \) or \( 62.5 \, \text{pm} \)

This wavelength is quite small, which is typical for particles at the atomic scale, and it reflects the wave-particle duality of electrons as described by quantum mechanics.