When an electron falls from $n = 5$ to $n = 2$ , the wavelength has been determined to be $434 n m$ .
What is the associated frequency?

6.91 \times 10^{- 14} H z

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6.91 \times 10^{- 5} H z

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6.91 \times 10^{5} H z

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6.91 \times 10^{14} H z

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Solution

The correct option is D $6.91 \times 10^{14} H z$
$ν λ = c \Rightarrow ν = \frac{c}{λ}$
$∴ ν = \frac{3 \times 10^{8}}{434 \times 10^{- 9}} = 6.91 \times 10^{14} ㎐$

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When an electron falls from n=5 to n=2, the wavelength has been determined to be 434 nm.What is the associated frequency?

The correct option is D 6.91×1014Hzνλ=c⇒ν=cλ∴ν=3×108434×10−9=6.91×1014㎐

When an electron falls from $n = 5$ to $n = 2$ , the wavelength has been determined to be $434 n m$ .
What is the associated frequency?

The correct option is D $6.91 \times 10^{14} H z$
$ν λ = c \Rightarrow ν = \frac{c}{λ}$
$∴ ν = \frac{3 \times 10^{8}}{434 \times 10^{- 9}} = 6.91 \times 10^{14} ㎐$