Question

energy of an Einstein of a radiation of wavelength 253.7 nm is
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Answer 1

An important reference wavelength is 253.7 nm, which corresponds to the main emission line in a low-pressure mercury lamp (used very often for the study of polymer photodegradation). The equivalent energy at this wavelength is 112.7 kcal mol-I.

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Answer 2

The energy of the radiation is 0.078×10-¹⁷J.

Given:-

The wavelength of the radiation = 253.7nm

To Find:-

The energy of the radiation.

Solution:-

We can easily find out the value of energy of the radiation by using these simple steps.

As

The wavelength of the radiation (l) = 253.7nm

Speed of light (c) = 3×10⁸m/s

Plank constant (h) = 6.6×10-³⁴

Energy of the radiation (E) =?

Here, first we have to convert all the units in their standard form as to put the values in the formula.

So,

The wavelength of the radiation (l) = 253.7×10-⁹m

According to the formula of Energy,

$E = \frac{hc}{l}$

$E = \frac{6.6 \times {10}^{ - 34} \times 3 \times {10}^{8} }{253.7 \times {10}^{ - 9} }$

$E = \frac{19.8 \times {10}^{ - 34 + 8} }{253.7 \times {10 }^{ - 9} }$

$E = \frac{0.078 \times {10}^{ - 26} }{10 - 9}$

$E = 0.078 \times {10}^{ - 26+ 9}$

$E = 0.078 \times {10}^{ - 17} j$

Hence, The energy of the radiation is 0.078×10-¹⁷J.

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