JEE MAIN - Chemistry (2024 - 8th April Evening Shift - No. 29)

Wavenumber for a radiation having 5800 $$\mathop A\limits^o $$ wavelength is $$x \times 10 \mathrm{~cm}^{-1}$$ The value of $$x$$ is ________. (Integer answer)
Answer
1724

Explanation

The wavenumber of a radiation is defined as the number of wavelengths per unit distance and is the reciprocal of the wavelength. Wavenumber is commonly represented in units of $$\mathrm{cm}^{-1}$$.

First, convert the given wavelength from angstroms ($$\mathop A\limits^o $$) to centimeters (cm). We know that:

$$1 \mathop A\limits^o = 10^{-8} \, \text{cm}$$

Given wavelength is 5800 $$\mathop A\limits^o $$:

$$5800 \mathop A\limits^o = 5800 \times 10^{-8} \, \text{cm}$$

Now calculate the wavenumber ($$\tilde{\nu}$$) which is the reciprocal of the wavelength:

$$\tilde{\nu} = \frac{1}{{5800 \times 10^{-8} \, \text{cm}}}$$

Simplify the expression:

$$\tilde{\nu} = \frac{1}{5800 \times 10^{-8} \, \text{cm}} = \frac{10^8}{5800} \, \mathrm{cm}^{-1}$$

Now, divide the numerator by the denominator to calculate the precise value:

$$\tilde{\nu} = \frac{10^8}{5800} \approx 1.724 \times 10^4 \, \mathrm{cm}^{-1}$$

Here, it is given that the wavenumber is in the form of $$x \times 10 \, \mathrm{cm}^{-1}$$, so $$x$$ would be the value we calculated divided by 10:

$$x = \frac{1.724 \times 10^4}{10} = 1724$$

Thus, the integer answer for the value of $$x$$ is:

1724

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