Question #1427c

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Question #1427c

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Answer 1 · 2017-11-16T22:18:04

$E_{γ} = 4.31 \cdot 10^{- 33} J = 26.94$ $E_(gamma)=4.31*10^(-33)J=26.94$ $f e V$ $feV$

Explanation:

$E_{γ} = h v$ $E_(gamma)=hv$
where:

$E_{γ}$ $E_(gamma)$ is the photon energy (in $J$ $J$ ).
$h$ $h$ is the Planck constant ( $6.63 \cdot 10^{- 34} J \cdot s$ $6.63*10^(-34)J*s$ ).
$v$ $v$ is the frequency (in $H z = \frac{1}{s}$ $Hz=1/s$ ).

So if you substitute with the values,
$E_{γ} = 6.63 \cdot 10^{- 34} \cdot 6.5$ $E_(gamma)=6.63*10^(-34)*6.5$
$E_{γ} = 4.31 \cdot 10^{- 33} J$ $E_(gamma)=4.31*10^(-33)J$

if you want it in $e V$ $eV$ and not in $J$ $J$ , you use the equivalence:
$1$ $1$ $e V = 1.6 \cdot 10^{- 19}$ $eV=1.6*10^(-19)$ $J$ $J$

so,
$E_{γ} = 2.694 \cdot 10^{- 14} e V$ $E_(gamma)=2.694*10^(-14)eV$

or,
$E_{γ} = 26.94$ $E_(gamma)=26.94$ $f e V$ $feV$

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Question #1427c

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