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प्रश्न
The quantum mechanical treatment of the hydrogen atom gives the energy value:
En = `(-13.6)/"n"^2` ev atom−1
- use this expression to find ΔE between n = 3 and n = 4
- Calculate the wavelength corresponding to the above transition.
उत्तर
En = `(-13.6)/"n"^2` ev atom−1
n = 3 E3 = `(-13.6)/3^2 = (-13.6)/9` = −1.51 ev atom−1
n = 4 E4 = `(-13.6)/4^2 = (-13.6)/16` = −0.85 ev atom−1
∆E = (E4 – E3) = (–0.85) – (–1.51) eV atom–1
= (–0.85 + 1.51)
= 0.66 eV atom–1
1eV = 1.6 × 10–19 J
∆E = 0.66 × 1.6 × 10–19 J
∆E = 1.06 × 10–19 J
hv = 1.06 × 10–19 J
`"hc"/λ` = 1.06 × 10–19 J
∴ λ = `"hc"/(1.06 xx 10^-19 "J")`
= `(6.626 xx 10^-34 "JS" xx 3 xx 10^8 "ms"^-1)/(1.06 xx 10^-19 "J")`
λ = 1.875 × 10–16 m
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