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Question: The radiation is emitted when a hydrogen atom goes from a higher energy state to a lower energy stat...

The radiation is emitted when a hydrogen atom goes from a higher energy state to a lower energy state. The wavelength of one line in the visible region of the atomic spectrum of hydrogen is 6.63×10−76.63 \times {10^{ - 7}} m.
Energy difference between the two state is:
A.3.0×10−19J3.0 \times {10^{ - 19}}J
B.2.98×10−19J2.98 \times {10^{ - 19}}J
C.5.0×10−19J5.0 \times {10^{ - 19}}J
D.6.5×10−19J6.5 \times {10^{ - 19}}J

Explanation

Solution

difference in energy is, ΔE=hϑ=hcλ\Delta E = h\vartheta = \dfrac{{hc}}{\lambda } where h = Planck’s constant

Complete step by step answer:
The difference in energy is ΔE\Delta E .
ΔE=hϑ=hcλ\Delta E = h\vartheta = \dfrac{{hc}}{\lambda } Here, h is the Planck’s constant
Where, ∆E is the energy in one quantum of electromagnetic radiation to the frequency of that radiation, h is the Planck’s constant, ϑ\vartheta is the frequency of electromagnetic radiation, and is its wavelength and c is the velocity of light.
⇒ΔE=(6.62×10−34)(3×108)6.63×10−7\Rightarrow \Delta E = \dfrac{{(6.62 \times {{10}^{ - 34}})(3 \times {{10}^8})}}{{6.63 \times {{10}^{ - 7}}}}
Where,
h=6.62×10−34h = 6.62 \times {10^{ - 34}} js
c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}}

⇒ΔE=2.98×10−19J\Rightarrow \Delta E = 2.98 \times {10^{ - 19}}J
So, the energy difference between the two states is ΔE=2.98×10−19J\Delta E = 2.98 \times {10^{ - 19}}J .

Therefore, the correct answer is option (B).

Note: When Electrically charged particles travel in acceleration, it generates an electromagnetic field also known as electromagnetic radiation. It is represented by different kinds of units. These are characterised by the properties named as frequency and wavelength. The SI unit of frequency is Hertz. It is also the number of waves which passes a given point in one second. Wavelength should be in the unit of length. Other quantities used is wave number which is defined as the number of wavelengths per unit length. Its unit is meter−mete{r^ - } . Electromagnetic waves in vacuum travel at the speed of 3×108m−2s3 \times {10^8}{m^{ - 2}}s which is also the velocity of the light. It is represented by the symbol c. c=ϑλc = \vartheta \lambda . Also, we can write it as λ=cϑ\lambda = \dfrac{c}{\vartheta } .