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4 years ago

Sub groups of metals

Chemistry

1 answer:

melamori03 [73]4 years ago

3 0

Answer:

Metalloids

Alkaline earth metals

Alkali

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3 years ago

In the emission spectrum of hydrogen the transitions observed in this experiment are in the visible region corresponding to the

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

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

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$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Higher energy level

$n_i$ = Lower energy level

1) The wavelength of the n=4 to n=1.

$n_i=1,n_f=4$

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{4^2} \right )\\\\\lambda =9.273\times 10^{-8}m=92.73 nm$

$1m=10^9nm$

The region of this electromagnetic transition will be in a ultraviolet region.

2) The wavelength of the n=4 to n=1.

$n_i=3,n_f=4$

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{4^2} \right )\\\\\lambda =1.875\times 10^{-6}m=1875 nm$

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The region of this electromagnetic transition will be in an infrared region.

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3 years ago