Question

An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum required for it to escape from the atom. What is the wavelength of the emitted electron?

Answer 1

To determine: The wavelength of the emitted electron of the hydrogen atom

Given Data: The energy absorbed by the electron in its ground state $= 1.5 \times$ The energy required for the electron to escape from the hydrogen atom.

Constants: Planck's constant, $h\quad =\quad 6.63\quad \times \quad { 10 }^{ -34 }\quad kg\sfrac { { m }^{ 2 } }{ s }$

Charge of an electron, $e\quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad C$

Mass of an electron $=\quad 9.1\quad \times \quad { 10 }^{ -31 }\quad kg$

Formulas to be used:

1) Ionisation energy of the electron = 13.6  

2) Kinetic energy of an electron in terms of potential, $V\quad =\quad e.V$

3) Wavelength, $\lambda \quad =\quad \frac { h }{ mu }$

Calculations:

Step 1:  Calculate the kinetic energy of the electron ejected out ${ E }_{ k }$

The incident energy, ${ I }_{ E }\quad =\quad 1.5\quad \times \quad Ionisation\quad Energy\quad of\quad the\quad electron\quad =\quad 1.5\quad \times \quad 13.6eV\quad$

${ E }_{ k }\quad =\quad 1.5\quad \times \quad 13.6eV\quad =\quad 6.8eV\quad =\quad 1.6\quad \times \quad { 10 }^{ -19 }\quad J$

Step 2: Determine u in terms of   using formula (3)

$\frac { 1 }{ 2 } m{ u }^{ 2 }\quad =\quad { E }_{ k }$

${ u }^{ 2 }\quad =\quad \frac { 2{ E }_{ k } }{ m }$

$u\quad =\quad \sqrt { \frac { 2{ E }_{ k } }{ m } }$

Step 3: Replace the u obtained from Step 2 in formula (2)  

$\lambda \quad =\quad \frac { h }{ mu } \quad =\quad \frac { h }{ m\sqrt { \frac { 2{ E }_{ k } }{ m } } } \quad =\quad \frac { h }{ \sqrt { 2m{ E }_{ k } } }$

Step 4: Substitute the known values to determine the wavelength of the emitted electron

$\lambda \quad =\quad \frac { h }{ mu } \quad =\quad \frac { 6.63\quad \times \quad { 10 }^{ -34 } }{ \sqrt { 2\quad \times \quad 9.1\quad \times \quad { 10 }^{ -31 }\quad \times \quad 6.8\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 } } } \quad =\quad \frac { 6.63\quad \times \quad { 10 }^{ -34 } }{ 14.07 } \quad =\quad 4.71\AA$