Question

An electron, in a hydrogen atom, in its ground state absorbs \[1.5\]times as much energy as the minimum required for its escape i.e.., \[13.6eV\] from the atom. Calculate the value of \[\lambda \] for the emitted electron.
A.\[4.69{A^0}\]
B.\[3.9{A^0}\]
C.\[7.9{A^0}\]
D.\[46.9{A^0}\]

Answer 1

Hint: The energy absorbed is the product of \[1.5\] times of \[13.6eV\] , as the minimum energy is \[13.6eV\] From this value of energy absorbed the kinetic energy will be determined. Substituting the kinetic energy value in the de broglie wavelength formula gives the value of wavelength.

Complete answer:
Given that an electron in a hydrogen atom absorbs \[1.5\]times as much energy as the minimum required for its escape i.e.., \[13.6eV\] from the atom
The amount of energy absorbed by an electron will be \[1.5 \times 13.6 = 20.4eV\]
Out of this \[20.4eV\] of energy \[20.4 - 13.6 = 6.8eV\] of energy will be converted into kinetic energy.
Convert this energy into joules which will be \[6.8 \times 1.6 \times {10^{ - 19}}\] joules
The equation for the kinetic energy is given as \[E = \dfrac{1}{2}m{v^2}\]
The velocity will be \[v = \sqrt {\dfrac{{2E}}{m}} \]
The de broglie wavelength will be given as \[\lambda = \dfrac{h}{{mv}}\]
Substitute the value of velocity in the above equation, will get \[\lambda = \dfrac{h}{{\sqrt {2Em} }}\]
\[h\] is Planck’s constant has the value of \[6.62 \times {10^{ - 34}}Js\]
\[m\] is mass of electron has the value of \[9.1 \times {10^{ - 31}}kg\]
Substitute the values in the wavelength formula
\[\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 6.8 \times 1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^{ - 31}}} }} = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{14.07}} = 4.69 \times {10^{ - 10}}m\]
While converts metres into angstroms, one angstrom will be equal to \[4.69{A^0}\]

Option A is the correct one.

Note:
The energy of the electron should be in column but not in electron volts, conversion must be made from electron volt to coulomb. The mass of the electron should be in kilograms. Finally, the obtained wavelength is in metres, it should be converted into angstroms. Thus, units must check while deriving the wavelength.