Question

An object absorbs energy corresponding to wavelength 2400 Å and emits two different radiations, the wavelength of one radiation 6000 Å. what is the wavelength of the other radiation?

Answer 1

The wavelength of the other radiation is 4000 Å.

Answer 2

The problem describes an object absorbing energy from a radiation of a given wavelength and then emitting energy as two different radiations with specified wavelengths. According to the principle of energy conservation, the total energy absorbed is equal to the sum of the energies of the emitted radiations.

The energy of a photon is given by the formula $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency. The frequency $\nu$ is related to the wavelength $\lambda$ and the speed of light $c$ by the equation $c = \lambda\nu$. Therefore, the energy can also be expressed as $E = \frac{hc}{\lambda}$.

Let $\lambda_{abs}$ be the wavelength of the absorbed radiation, and $E_{abs}$ be the corresponding energy.
Let $\lambda_1$ and $\lambda_2$ be the wavelengths of the two emitted radiations, and $E_1$ and $E_2$ be their corresponding energies.

According to the conservation of energy:
$E_{abs} = E_1 + E_2$

Substituting the formula for energy in terms of wavelength:
$\frac{hc}{\lambda_{abs}} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$

Dividing both sides by $hc$ (since $hc$ is a non-zero constant):
$\frac{1}{\lambda_{abs}} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$

We are given:
Wavelength of absorbed radiation, $\lambda_{abs} = 2400 \, Å$.
Wavelength of one emitted radiation, $\lambda_1 = 6000 \, Å$.
We need to find the wavelength of the other emitted radiation, $\lambda_2$.

Substitute the given values into the equation:
$\frac{1}{2400 \, Å} = \frac{1}{6000 \, Å} + \frac{1}{\lambda_2}$

To find $\lambda_2$, we rearrange the equation:
$\frac{1}{\lambda_2} = \frac{1}{2400 \, Å} - \frac{1}{6000 \, Å}$

To subtract the fractions on the right side, we find a common denominator for 2400 and 6000. The least common multiple of 2400 and 6000 is 12000.
We convert the fractions to have the denominator 12000:
$\frac{1}{2400} = \frac{1 \times 5}{2400 \times 5} = \frac{5}{12000}$
$\frac{1}{6000} = \frac{1 \times 2}{6000 \times 2} = \frac{2}{12000}$

Now, substitute these back into the equation:
$\frac{1}{\lambda_2} = \frac{5}{12000 \, Å} - \frac{2}{12000 \, Å}$
$\frac{1}{\lambda_2} = \frac{5 - 2}{12000 \, Å}$
$\frac{1}{\lambda_2} = \frac{3}{12000 \, Å}$

Simplify the fraction on the right side:
$\frac{1}{\lambda_2} = \frac{1}{4000 \, Å}$

Taking the reciprocal of both sides gives the value of $\lambda_2$:
$\lambda_2 = 4000 \, Å$

Thus, the wavelength of the other radiation is 4000 Å.

Explanation: The energy absorbed by the object is equal to the sum of the energies of the two emitted radiations, based on energy conservation. Using the relationship $E = hc/\lambda$, the conservation of energy translates to $\frac{hc}{\lambda_{abs}} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$. Dividing by $hc$, we get $\frac{1}{\lambda_{abs}} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$. Substituting the given values $\lambda_{abs} = 2400 \, Å$ and $\lambda_1 = 6000 \, Å$, we solve for $\lambda_2$: $\frac{1}{\lambda_2} = \frac{1}{2400} - \frac{1}{6000} = \frac{5}{12000} - \frac{2}{12000} = \frac{3}{12000} = \frac{1}{4000}$. Therefore, $\lambda_2 = 4000 \, Å$.