Question

The energy required to melt $1$g ice is $33J$. The number of quanta of radiation of frequency $4.67\times {10}^{13}se{c}^{-1}$ that must be absorbed to melt $10$g ice is:

• A. $1.065\times {10}^{22}$
• B. $3.205\times {10}^{23}$
• C. $9.076\times {10}^{20}$
• D. none of the above

Answer 1

The correct option is A $1.065\times {10}^{22}$

Let the energy of the radiation

$E=hv$

Given that:

$h=6.625\times {10}^{-34}J.s$

$v=4.67\times {10}^{13}{\sec }^{-1}$

$E=6.626\times {10}^{-34}J.s\times 4.67\times {10}^{13}{\sec }^{-1}$

$=3.09\times {10}^{-20}J$

Let the melt $10g$ ice is $330$$J$

Number of quanta $=\frac{330}{3.09\times {10}^{-20}}$

$=1.065\times {10}^{22}$