An excited electron in $n = 5$ emits enough energy to fall to $n = 2$ .
What is the wavelength associated with this energy emission?

343 n m

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434 n m

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565 n m

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646 n m

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Solution

The correct option is A $343 n m$
Let the wavelength be $λ$ . Thus by Rydberg's formula
$\frac{1}{λ} = R (\frac{1}{{(2)}^{2}} - \frac{1}{{(5)}^{2}}) = \frac{21 R}{100}$
$∴ λ = \frac{100}{21 R} = 434 ㎚$
Option A

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An excited electron in n=5 emits enough energy to fall to n=2.What is the wavelength associated with this energy emission?

The correct option is A 343 nmLet the wavelength be λ. Thus by Rydberg's formula1λ=R(1(2)2−1(5)2)=21R100∴λ=10021R=434㎚Option A

An excited electron in $n = 5$ emits enough energy to fall to $n = 2$ .
What is the wavelength associated with this energy emission?

The correct option is A $343 n m$
Let the wavelength be $λ$ . Thus by Rydberg's formula
$\frac{1}{λ} = R (\frac{1}{{(2)}^{2}} - \frac{1}{{(5)}^{2}}) = \frac{21 R}{100}$
$∴ λ = \frac{100}{21 R} = 434 ㎚$
Option A