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Q.

As energy associated with changes with its wavelength, often the reciprocal of the wavelength $1 / λ$ . is used to describe energy associated with that wavelength. Then, mark the correct equivalence.

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a

$1 eV = 5092.6 {cm}^{- 1} and 1 {cm}^{- 1} = 1.2398 \times 10^{- 4} eV$

b

$1 eV = 1239.8 {cm}^{- 1} and 1 {cm}^{- 1} = 8068 \times 10^{- 7} eV$

c

$1 eV = 8068.8 {cm}^{- 1} and 1 {cm}^{- 1} = 6.65 \times 10^{- 19} eV$

d

$1 eV = 8065 \times 10^{- 19} {cm}^{- 1} and 1 {cm}^{- 1} = 1.2398 \times 10^{- 4} eV$

answer is C.

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Detailed Solution

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$\begin{array}{l} As, h f = E \\ \Rightarrow \frac{h c}{λ} = E \Rightarrow \frac{1}{λ} = \frac{E}{h c} \\ Now, h c \approx 1240 eV - nm \\ \Rightarrow h c \approx 1240 \times 10^{- 7} eV - cm \\ So, \frac{1}{λ} = \frac{1 eV}{1240 \times 10^{- 7} eV - cm} (when E = 1 eV) \\ = 8065.8 {cm}^{- 1} \end{array}$

So, a wave length reciprocal $\frac{1}{λ}$ of 8065.8 cm^-1 corresponds to an energy of 1 eV.

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As energy associated with changes with its wavelength, often the reciprocal of the wavelength 1/λ. is used to describe energy associated with that wavelength. Then, mark the correct equivalence.