Let's substitute the value of hh into the Planck equation E = h fE=hf:
Rightarrow E = (4.136 times 10^(- 15)⇒E=(4.136×10−15 "e V"e V cdot⋅ "s")s) times f×f
Then, let's substitute the definition of ff, f = frac(c)(lambda)f=cλ, into the above equation:
Rightarrow E = (4.136 times 10^(- 15)⇒E=(4.136×10−15 "e V"e V cdot⋅ "s")s) times frac(c)(lambda)×cλ
Let's substitute the values of cc and lambdaλ into the equation:
Rightarrow E = (4.136 times 10^(- 15)⇒E=(4.136×10−15 "e V"e V cdot⋅ "s")s) times frac(3.00 times 10^(8) " m s"^(- 1))(522 " nm")×3.00×108 m s−1522 nm
Rightarrow E = (4.136 times 10^(- 15)⇒E=(4.136×10−15 "e V"e V cdot⋅ "s")s) times frac(3.00 times 10^(8) " m s"^(- 1))(5.22 times 10^(- 7) " m")×3.00×108 m s−15.22×10−7 m
Rightarrow E = frac(1.2408 times 10^(- 6) " e V" cdot "m")(5.22 times 10^(- 7) " m")⇒E=1.2408×10−6 e V⋅m5.22×10−7 m
Rightarrow E = 2.337⇒E=2.337 "e V"e V
therefore E = 2.38 "e V"
Therefore, this band corresponds to a decrease of 2.38 "e V".