7. The sodium flame test has a characteristic
yellow colour due to the emission of wavelength
of 500 nm. The mass equivalence of one photon
of this wavelength is
(a) 3.7 * 10-26 kg
(b) 5.7 x 10-30 kg
(c) 4.4 x 10-36 kg
(d) 9.7 x 10-40 kg
Answers
Answer :
3.75×10^−36kg
Solution :
λ
λ=
λ=h
λ=hm
λ=hmv
λ=hmvor
λ=hmvorm
λ=hmvorm=
λ=hmvorm=h
λ=hmvorm=hλ
λ=hmvorm=hλ×
λ=hmvorm=hλ×v
λ=hmvorm=hλ×v=
λ=hmvorm=hλ×v=6.626
λ=hmvorm=hλ×v=6.626×
λ=hmvorm=hλ×v=6.626×10
λ=hmvorm=hλ×v=6.626×10−
λ=hmvorm=hλ×v=6.626×10−34
λ=hmvorm=hλ×v=6.626×10−34k
λ=hmvorm=hλ×v=6.626×10−34kg
λ=hmvorm=hλ×v=6.626×10−34kgm
λ=hmvorm=hλ×v=6.626×10−34kgm2
λ=hmvorm=hλ×v=6.626×10−34kgm2s
λ=hmvorm=hλ×v=6.626×10−34kgm2s−
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×10
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108m
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×10
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×10−
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×10−36
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×10−36k
λ=hmvorm=hλ×v=6.626×10−34kgm2s−1(589×10−9)(3×108ms−1)=3.75×10−36kg
Answer:
correct answer is option C
4.4×10^36 kg