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 $2.38$ $"e V"$.