Question

Iodine molecule is dissociates into atoms after absorbing light of 4500 A°. If one quantum of radiation is absorbed by each molecule.calculate K.E of I-atoms(B.E of I2=240 kJ/mole)​

Answer 1

Answer:

$\huge{\boxed{\sf{K.E\:OF\:I-ATOM=0.21\times {10}^{-19}}}}$

Explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$B.E \: of \: i_{2} = 240 \frac{KJ}{mole} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 240 \times {10}^{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240 \times {10}^{3} }{na} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240 \times {10}^{3} }{6.022 \times {10}^{23} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 39.85 \times {10}^{ - 20} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3.985 \times {10}^{ - 19} \\ again \\ energy \: of \: photon = \frac{hc}{ \lambda } = \frac{12400 \: e.v }{4500} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2.755 \times 1.6 \times {10}^{ - 19} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4.408 \times {10}^{ - 19} j \\ extra \: energy = (4.408 - 3.985) \times {10}^{ - 19} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.423 \times {10}^{ - 19} j \\ extra \: energy \: converted \: into \: K.E \: of \: two \: iodine = 0.21 \times {10}^{ - 19} J$

Answer 2

Solution:

Given:

=> B.E of I2=240 kJ/mole

So,

$\sf{\implies Energy\;provided\;to\;iodine\;molecule = \dfrac{hc}{\lambda}}$

$\sf{\implies \dfrac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4500 \times 10^{-10}}}$

$\sf{\implies 4.417 \times 10^{-19}J}$

$\sf{Also,\;energy\;used\;for\;breaking\;up\;iodine\;molecule=\dfrac{240\times 10^{3}}{6.023 \times 10^{23}}}$

$\sf{\implies 3.984\times 10^{-19}J}$

$\sf{Energy\;used\;for\;providing\;Kinetic\;energy\;to\;two\;I\;atoms =[4.417-3.984]\times 10^{-19}J}$

$\sf{So,\;K.E\;per\;iodine\;atom = \bigg[\dfrac{4.417-3.984}{2}\bigg]\times 10^{-19}}$

${\boxed{\boxed{\sf{\implies 2.16\times 10^{-20}J}}}}$