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2 years ago

A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?

Chemistry

1 answer:

Rama09 [41]2 years ago

4 0

The frequency of the radiation is equal to $7.17 \times 10^{19}$ Hertz.

Given the following data:

Photon energy = $4.75 \times 10^{-14}$ Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

Where:

h is Planck constant.

f is photon frequency.

Substituting the given parameters into the formula, we have;

$4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}$

Frequency, F = $7.17 \times 10^{19}$ Hertz

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Answer: The reaction proceeds in the forward direction

Explanation:

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Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

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$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

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