Question

7. A single pulse of a laser yields an average of 5.00 x 1018 photons with = 532 nm. If melting ice to water at 0°C requires 6.01 kJ/mol, what is the fewest number of laser pulses needed to melt 125.3 g of ice?

Answer 1

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 125.3 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(1.253*10^2 g)/(18.02 g/mol)

= 6.955 mol

Heat required,

Q = number of mol * Hfus

= 6.955 mol * 6.01 KJ/mol

= 41.800 KJ

= 41800 J

Now calculate the number of photons.

Given:

lambda = 5.32*10^-7 m

1st calculate energy of 1 photon

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(5.32*10^-7 m)

= 3.736*10^-19 J

number of photon = total energy/energy of 1 photon

n = 41800.0/3.736*10^-19

= 1.119*10^23

Number of pulses = number of photon / number of photons in a pulse

= 1.119*10^23 / 5.00*10^18

= 2.24*10^4 pulses

Answer: 2.24*10^4 pulses