Question

The energy levels of an atom is shown in figure:

 Which one of these transitions will result in the emission of a photon of wavelength $1241\, nm$ ? Given $\left( h =6.62 \times 10^{-34} Js \right)$

• A. D
• B. B
• C. A
• D. $C$

Answer 1

The Correct Option is A

The correct answer is (A) : D
$\lambda =\frac{hc}{\Delta E}$
$\Delta E_A=2.2eV$
$\Delta E_B=5.2eV$
$\Delta E_C=3eV$
$\Delta E_D=10eV$
${\lambda }_A=\frac{6.62\times 10^{-34}\times 3\times 10^8}{2.2\times 1.6\times 10^{-19}}$
$=\frac{12.41\times 10^{-7}}{2.2}m$
$=\frac{1241}{2.2}nm=564nm$
${\lambda }_B=\frac{1241}{5.2}nm=238.65nm$
${\lambda }_C=\frac{1241}{3}nm=413.66nm$
${\lambda }_D=\frac{1241}{10}=124.1nm$

Answer 2

Each transition's energy difference is associated with a particular light wavelength. One can compute the wavelength \( \lambda \) of the photon that was released. Using the formula: \[ \lambda = \frac{hc}{\Delta E} \] where \( h \) is Planck's constant,
\( c \) is the speed of light,
and \( \Delta E \) is the energy difference.
Calculating \( \Delta E \) for each transition:
$\Delta E_A = 2.2 \text{ eV}$
$\Delta E_B = 5.2 \text{ eV}$
$\Delta E_C = 3 \text{ eV}$
$\Delta E_D = 10 \text{ eV}$
$\lambda_A = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}}$
$= \frac{12.41 \times 10^{-7}}{2.2} \text{ m}$
$= \frac{1241}{2.2} \text{ nm} = 564 \text{ nm}$
$\lambda_B = \frac{1241}{5.2} \text{ nm} = 238.65 \text{ nm}$
$\lambda_C = \frac{1241}{3} \text{ nm} = 413.66 \text{ nm}$
$\lambda_D = \frac{1241}{10} = 124.1 \text{ nm}$ This matches the given wavelength for transition \( D \).