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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 47 of 67
Marks: +1, -0
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s, (c) energy of quantum and(d) number of quanta present if it produces 2 J of energy.
Solution:  
Wavelength of light (λ) = 616 nm = 616 × 10−910^{-9} m [Since 1 nm = 10−910^{-9} m]
(a) Frequency of emission (v) = cλ\frac{c}{\lambda} = 3×108616×10−9\frac{3 \times 10^8}{616 \times 10^{-9}} = 4.87 × 1014 sec−110^{14}\,\text{sec}^{-1}
(b) Distance travelled in 30 sec = 30 × 616 × 10−910^{-9} × 4.87 × 101410^{14} = 9.0 × 10910^9 m
(c) Energy of quanta (E) = hv = 6.626 × 10−3410^{-34} × 4.87 × 101410^{14} = 3.227 × 10−1910^{-19} J
(d) Number of quanta in 2 J of energy = 23.227×10−19\frac{2}{3.227 \times 10^{-19}} = 6.2 × 101810^{18} photons
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