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natita [175]
3 years ago
5

An electron undergoes a transition from an initial (ni) to a final (nf ) energy state. The energies of the ni and nf energy stat

es are −2.179×10−18J and −8.720×10−20 J, respectively.. . Calculate the wavelength (λ) of the light in nanometers (nm) corresponding to the energy change (ΔE) value of this transition. You can use the following values for your calculations:. . Planck′s constant (h) = 6.626×10−34 J⋅s. speed of light (c) = 2.998×108 m/s. 1 m = 109 nm
Physics
1 answer:
Ipatiy [6.2K]3 years ago
6 0
The equation for the energy (E) of the electron may be obtained by the equation,
                                             E = hc / λ
where h is Planck's constant, c and λ are speed of light and wavelength, respectively. Substituting the values,
              (-2.179x10^-18 - -8.720x10^-20) = (6.626x10^-34)(2.998x10^8)/ λ
From the equation, the value of λ is approximately equal to 9.496x10^-8 m. 
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4 years ago
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Answer:

649.8420 nm

Explanation:

We have given \Theta _1=24^{\circ} \ and\ \Theta _2=39^{\circ}

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The condition for constructive interference is dsin\Theta =n\lambda

Where d is distance between slits \Theta is angle of emergence and n is order of maxima

Using constructive interference equation

dsin\Theta _1=n\lambda _1 ---------eqn 1

dsin\Theta _2=n\lambda _2---------eqn 2

On dividing eqn 1 by eqn 2

\frac{\lambda _1}{\lambda _2}=\frac{sin\Theta _1}{sin\Theta _2}

\frac{420}{\lambda _2}=\frac{sin24^{\circ}}{sin39^{\circ}}

\lambda _2=649.8420\ nm  

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3 years ago
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