By using photons of specific wavelengths, chemists can dissociate gaseous HI to produce H atoms with certain speeds. When HI dissociates, the H atoms move away rapidly, whereas the heavier I atoms move more slowly. If a photon of 231 nm is used, what is the excess energy (in J) over that needed for dissociation

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Explanation:

Given the data in the question;

wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m

we determine the energy of the proton;

E = hc / λ

where h is plank constant ( 6.626 × 10⁻³⁴ JS )

and c is the speed of light ( 3 × 10⁸ m/s )

we substitute

E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]

E = 8.61 × 10⁻¹⁹ J

we know that, bond energy for H-I is 295 kJ/mol

so, H = 295 × 10³ J/mol

Now, energy to dissociate HI will be;

⇒ H / N

where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )

energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )

= 4.898 × 10⁻¹⁹ J

Therefore, Excess energy over dissociation will be;

⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )

= 3.712 × 10⁻¹⁹ J

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

answer is a

explanation: