Question

if $n = 1=2.62\times10^{-21}\text{ j}$
$n = 2 = 3.75\times10^{-21}\text{ j}$
$n = 3 = 4.63\times10^{-21}\text{ j}$
calculate the energy of $2.95\times10^{25}$ photons
shown by the arrow

Explanation:

Step1: Determine energy difference

Find the energy difference between $n = 3$ and $n = 1$.
$\Delta E=E_3 - E_1$
$E_3 = 4.63\times10^{-21}\text{ J}$, $E_1 = 2.62\times10^{-21}\text{ J}$
$\Delta E=(4.63\times 10^{-21}-2.62\times 10^{-21})\text{ J}=2.01\times 10^{-21}\text{ J}$

Step2: Calculate total energy for given number of photons

Multiply the energy - difference by the number of photons.
The number of photons $N = 2.95\times10^{25}$
$E_{total}=N\times\Delta E$
$E_{total}=(2.95\times 10^{25})\times(2.01\times 10^{-21}\text{ J})$
Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $E_{total}=(2.95\times2.01)\times10^{25+( - 21)}\text{ J}$
$E_{total}=5.9295\times10^{4}\text{ J}$

Answer:

$5.93\times 10^{4}\text{ J}$ (rounded to two decimal - places)