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3 years ago

The sodium atom loses an electron to form a sodium ion (Na+). Which statement is correct with respect to its atomic radius?

2 answers:

6 0

The atomic number of sodium is 11. Sodium is a metal and it has the electronic configuration: 2, 8,1. This configuration implies that, the electrons in an atom of sodium are distributed into three different shells and the outermost shell has one electron in it. If sodium decide to give away the electron on its outermost shell, it will have 10 electrons left and those electrons will be distributed in only the first two shells and then it will be described as an ion. The third shell will not exist again because the electron there has been given away. Thus, the sodium ion is going to have a smaller atomic radius because its size has been reduced. This implies that the sodium ion will have a smaller radius than the sodium atom.

mixer [17]3 years ago

3 0

Its atomic radius increases because there is less electron shielding after the sodium loses one e-. Sodium is a nonmetal which means it forms an anion. Anions always increase from their original atomic radius

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boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

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$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

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~Jurgen

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