If a photon of 254 nm is used, the excess energy (in J) over that needed for dissociation is 2.93 10^-19 J. If this excess energy is carried away by the H atom as kinetic energy, then the speed is equal to the 2.4× 10^(-8) m/s..
<h3>What is Dissociation energy? </h3>
The bond-dissociation energy is defined as measure of the strength of a chemical bond which is represented as A−B. It can also be defined as the standard enthalpy change when the bond A−B is cleaved by homolysis to give two fragments A and B, that are usually radical species.
Energy of photon with λ 254 nm is given as
E = hC \ λ
= 6.626 × 10-³⁴ JS × 3 × 10 8 ms -1 /254 × 10 -9 m
= 7.83 10^-19 J
As we know that,
dissociation energy of HI = 295 10³ J. mol-¹
= 295 × 10³ J / mol -¹ / 6.023 × 10²³ mol -¹
= 4.898 × 10^-19 J.
Hence excess energy over dissociation
= 7.83 × 10^-19 J – 4.448 × 10^-19 J
= 2.93 × 10^-19 J
K.E = 2.93 × 10^-19 J
As we know that,
K.E = 1/2 of mv^2
mass of hydrogen atom = 1g = 10^(-3) kg.
2 × 2.93 × 10^-16 J = mv^2
v^2 = 5.86 × 10^(-16)
v = 2.42 × 10^(-8) m/s.
Thus, we concluded that the speed of H atom is
2.4× 10^(-8) m/s.
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