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Let's substitute the value of $h$ into the Planck equation $E = h f$:

$Rightarrow E = (4.136 times 10^(- 15)$ $"e V"$ $cdot$ $"s")$ $times f$

Then, let's substitute the definition of $f$, $f = frac(c)(lambda)$, into the above equation:

$Rightarrow E = (4.136 times 10^(- 15)$ $"e V"$ $cdot$ $"s")$ $times frac(c)(lambda)$

Let's substitute the values of $c$ and $lambda$ into the equation:

$Rightarrow E = (4.136 times 10^(- 15)$ $"e V"$ $cdot$ $"s")$ $times frac(3.00 times 10^(8) " m s"^(- 1))(522 " nm")$

$Rightarrow E = (4.136 times 10^(- 15)$ $"e V"$ $cdot$ $"s")$ $times frac(3.00 times 10^(8) " m s"^(- 1))(5.22 times 10^(- 7) " m")$

$Rightarrow E = frac(1.2408 times 10^(- 6) " e V" cdot "m")(5.22 times 10^(- 7) " m")$

$Rightarrow E = 2.337$ $"e V"$

$therefore E = 2.38$ $"e V"$

Therefore, this band corresponds to a decrease of $2.38$ $"e V"$.

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