**Calculate the minimum frequency of photon that can break a Hydrogen bond.****Calculate the maximum wavelength of a photon that can break a Hydrogen bond.**

The question aims to find the **minimum frequency** of a **photon** and its **maximum wavelength** that can break a **Hydrogen Bond** of a **protein molecule.**

The concepts needed to solve this problem include **Planck’s Equation** and **photon’s** (the smallest particle or packet of light) **frequency** using **Planck’s equation.** The equation is given as:

\[ E = h v \]

It can also be written as:

\[ E = h \dfrac{ c } { \lambda } \]

## Expert Answer

**a)** The **energy** of the **photon** is given as:

\[ E = 0.1 eV \]

To calculate the correct value, we need to convert the unit of **energy** from $eV$ to $J (Joules)$. It is given as:

\[ 1 eV = 1.6 \times 10^ {-19} J \]

\[ 0.1 eV \times 1 eV = 0.1 \times 1.6 \times 10^ {-19} J \]

\[ 0.1 eV = 1.6 \times 10^ { -20 } J \]

We can use **Planck’s Equation** to calculate the **frequency** of the **photon,** which is given as:

\[ E = h v \]

Here, $v$ is **frequency** of the **photon,** $E$ is the **energy** of the **photon,** and $h$ is **Planck’s constant.** The value of the Planck’ constant is given as:

\[ h = 6.626 \times 10^ { -34 } Js \]

Rearranging the formula to calculate the **frequency** of the **photon** is given as:

\[ v = \dfrac{ E }{ h } \]

Substituting the values in the given formula, we get:

\[ v = \dfrac{ 1.6 \times 10^ { -20 } J }{ 6.626 \times 10^ { -34 } Js } \]

Solving the equation, we get:

\[ v = 2.4 \times 10^ {13} Hz \]

**b)** To calculate the **wavelength** of the **photon,** we use the other form of the equation where the **frequency** is replaced by the **speed** of **light** and **wavelength** of the **light.** The equation is given as:

\[ E = h (\dfrac{ c }{ \lambda }) \]

The speed of light is given as:

\[ c = 3 \times 10^ { 8 } m/s \]

Rearranging the formula to calculate the **wavelength** of the **photon** as:

\[ \lambda = \dfrac{ hc }{ E } \]

Substituting the values, we get:

\[\lambda = \dfrac{ (6.626 \times 10^ { -34 } Js) . (3 \times 10^ { 8 } m/s) }{ 1.6 \times 10^ { -20} J }

Solving the equation, we get:

\[ \lambda = 1.24 \times 10^ { -5 } m \]

## Numerical Result

**a)** The **minimum frequency** of the **photon** required to break a **hydrogen bond** in a **protein molecule** while the energy of the photon is $0.1 eV$ is calculated to be:

\[ v = 2.4 \times 10^ { 13 } Hz \]

b) The **maximum wavelength** of the **photon** to break a **hydrogen bond** in a **protein molecule** while the energy of the photon is $0.1 eV$ is calculated to be:

\[ \lambda = 1.24 \times 10^ { -5 } m \]

## Example

Find the **frequency** of the **photon** with an **energy** of $5.13 eV$, which is required to break an **oxygen bond** in $O_2$.

The formula is given as:

\[ v = \dfrac{E}{h} \]

\[ v = \dfrac{5.13 \times 1.6 \times 10^{-19} J}{6.626 \times 10^{-34} Js}\]

\[ v = 1.24 \times 10^{15} Hz \]