Answer:
a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Explanation:
E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period
a) Period = 1 fs = 1 × 10⁻¹⁵ s
f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s
f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) Period = 1s
f = 1/period = 1.0 Hz
E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Answer:
(i) E = 6.626 X 10⁻¹⁹ J = 400 KJ/mol
(ii) E = 6.626 X 10⁻²⁰ J = 40 KJ/mol
(iii) E = 6.626 X 10⁻³⁴ J = 4 X 10⁻¹³ KJ/mol
Explanation:
Energy associated with excitation of a quantum is given as;
E = hf
where;
E is the energy of excitation
h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹
f is the threshold frequency in s⁻¹
Thus, E = h/t
Part (i)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)
E = 6.626 X 10⁻¹⁹ J
In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol
Part (ii)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)
E = 6.626 X 10⁻²⁰ J
In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol
Part (iii)
E = (6.626 X 10⁻³⁴)/(1)
E = 6.626 X 10⁻³⁴ J
In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol
