A difference of 2.3 eV separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is

Question

A difference of 2.3 eV separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is (A) 6.7× 1014Hz (B) 4.6× 1014Hz (C) 5.6 × 1014 Hz (D) 7.7× 1014Hz

Tardigrade Admin · Accepted Answer

Given, the difference in energy level E = 2.3 eV = 2.3 × 1.6 × 10- 19 J âPlanck's constant h = 6.63 × 10- 34 J s Let v be the frequency, then E = hf or f=(E/h)=( text2.3 × text1.6 × text10- text19/ text6.63 × text10- text34) = 5.6 × 1014 Hz

Q. A difference of $2.3 e V$ separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is

$6.7 \times 1 0^{14} Hz$

$4.6 \times 1 0^{14} Hz$

$5.6 \times 1 0^{14} Hz$

$7.7 \times 1 0^{14} Hz$

Given, the difference in energy level $E = 2.3 e V$
$= 2.3 \times 1.6 \times 1 0^{- 19} J$
Planck's constant $h = 6.63 \times 1 0^{- 34} J s$
Let $v$ be the frequency, then $E = h f$
or $f = \frac{E}{h} = \frac{2.3 \times 1.6 \times 10 ^{- 19}}{6.63 \times 10 ^{- 34}}$
$= 5.6 \times 1 0^{14} Hz$

Q. A difference of 2.3eV separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is

6.7×1014Hz

4.6×1014Hz

5.6×1014Hz

7.7×1014Hz