Bond energy of F 2 is 150 KJ mole -1- The minimum frequency of photon required to break F - F bond isA- 3 X 1014 HzB- 3-76 - 1014 HzC- 2-76 - 1014 HzD- 1-38 - 1014 Hz

The correct option is B 3.76 X 1014 Hz N × hv = 150 × 10^3 J v = 150 × 10^3/6.023 × 10^23 × 6.625 × 10^ 34 = 150 × 10^14/6.023 × 6.625 = 3.76 × 10^14 Hz ...

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Standard XII

Chemistry

1900's Physics and Wave

Bond energy o...

Question

Bond energy of $F_{2}$ is $150 K J m o l e^{- 1}$ . The minimum frequency of photon required to break F - F bond is

3 X 10¹⁴ Hz

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3.76 X 10¹⁴ Hz

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2.76 X 10¹⁴ Hz

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1.38 X 10¹⁴ Hz

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Solution

The correct option is B
3.76 X 10¹⁴ Hz

$N \times h v = 150 \times 10^{3} J$

$v = \frac{150 \times 10^{3}}{6.023 \times 10^{23} \times 6.625 \times 10^{- 34}}$

= $\frac{150 \times 10^{14}}{6.023 \times 6.625} = 3.76 \times 10^{14} H z$

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Bond energy of F2 is 150 KJ mole−1. The minimum frequency of photon required to break F - F bond is

The correct option is B 3.76 X 1014 Hz N × hv = 150 × 103 J v = 150 × 1036.023 × 1023 × 6.625 × 10−34 = 150 × 10146.023 × 6.625 = 3.76 × 1014 Hz

Bond energy of $F_{2}$ is $150 K J m o l e^{- 1}$ . The minimum frequency of photon required to break F - F bond is

The correct option is B
3.76 X 10¹⁴ Hz

$N \times h v = 150 \times 10^{3} J$

$v = \frac{150 \times 10^{3}}{6.023 \times 10^{23} \times 6.625 \times 10^{- 34}}$

= $\frac{150 \times 10^{14}}{6.023 \times 6.625} = 3.76 \times 10^{14} H z$