SATHEE: Structure of Atom - Result Question 2

Structure of Atom - Result Question 2

####2. According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as $(h = 6.62 \times 10^{- 27} .$ ergs, $c = 3 \times 10^{10} c m s^{- 1}$ , $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ )

[NEET Kar. 2013]

(a) $\frac{1.196 \times 10^{16}}{λ}$

(b) $\frac{1.196 \times 10^{8}}{λ}$

(d) $\frac{2.859 \times 10^{16}}{λ}$

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Solution:

(b) $E = \frac{h c}{λ} \times N_{A}$

$= \frac{6.62 \times 10^{- 27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{λ}$

$= \frac{1.19 \times 10^{8}}{λ} e r g s m o l^{- 1}$

Law of photochemical equivalence is the fundamental principle relating to chemical reactions induced by light which states that for every mole of a substance that reacts $6.022 \times 10^{23}$ quanta of light are absorbed. The photochemical equivalence law is also sometimes called the stark-Einstein law.

Structure of Atom - Result Question 2

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Structure of Atom - Result Question 2

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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