Question

In the ground state of hydrogen atom, electron absorbs 1.5 times energy than the minimum energy \( (2.18 \times 10^{-18} J)\) to escape from the atom. The wavelength of the emitted electron (in m) is \((m_e = 9 \times 10^{-31} kg)\)

• A. \( \frac{h \times 10^{24}}{\sqrt{1.962}} \)
• B. \( \frac{h}{\sqrt{1.962}} \times 10^{23} \)
• C. \( \frac{h \times 10^{25}}{\sqrt{1.962}} \)
• D. \( \frac{h}{\sqrt{1.962}} \times 10^{22} \)

Hint:

Remember to use the de Broglie wavelength formula \( \lambda = \frac{h}{p} \), where momentum \( p = \sqrt{2mE} \).

Answer 1

The Correct Option is A

Step 1: The energy required for the electron to escape is given as: \[ E = 2.18 \times 10^{-18} \, \text{J} \times 1.5 = 3.27 \times 10^{-18} \, \text{J} \] Step 2: The wavelength of the emitted electron can be found using the de Broglie equation: \[ \lambda = \frac{h}{p} \] where \( p = \sqrt{2mE} \), and \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( E \) is the energy. 

Step 3: Substituting values: \[ \lambda = \frac{h}{\sqrt{2m \times 3.27 \times 10^{-18}}} \]