What is the pH when [H+] = 0.000032 M?
1 answer:
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Answer:change in energy of the electron=3.052x 10^-19J
which energy level does it move= level 2 , n=2
Explanation:
Using the formulae
1/λ = R (1/n1²- 1/n2²)
Where λ= 656.27 nm
1 nm = 1 x 10^-9 m
656.27 nm = 656.27 x 1 x 10^-9 =6.5726 x 10^-7
R =Rydberg constant = 1.0967 x 10^7m-1
1/λ = R (1/n1²- 1/n2²)
1/6.5726 x 10^-7=1.0967 x 10^7(1/n1²- 1/3²)
1/n1²=(1/6.5726 x 10^-7 x 1/1.0967 x 10^7) + 1/9
1/n1²=1,521,467.9 x 9.118x10^-8 + 0.1111
1/n1² =0.2498
n1²= 1/0.2498 =4
n1= = 2
it moves to energy level 2
b) Change in energy =ΔE = Rhc (1/n1²- 1/n2²)
Where R==Rydberg constant = 1.0967 x 10^7m-1
h = Planck constant = 6.626x 10^-34js
c = speed of light = 3.0 x 10^8 x m/s
ΔE = Rhc (1/n1²- 1/n2²)
=1.0967 x 10^7m-1 x6.626x 10^-34js X 3.0 x 10^8 x m/s (1/2² - 1/3²)
=2.18 x 10-18 x ( 1 /4 - 1/9)
=3.052x 10^-19J