Question

When electronic transition occurs from an higher energy to a lower energy state with energy difference equal to expressed in electron volts the wavelength of line emitted is approximately equal to (1) \(\frac{12375}{\Delta E} \wedge\) (2) \(\frac{12375}{\Delta E} \times 10^{-8} \mathrm{~cm}\) (3) \(\frac{12375}{\Delta E} \times 10^{-10} \mathrm{~m}\) (4) Either of these

Short answer

The correct answer is (1) based on the formula and unit approximation

Step-by-step solution

Understand the Relationship

Recognize that the problem asks for the wavelength of light emitted when an electron transitions between energy states. The wavelength \[\lambda\] can be related to energy difference \[\Delta E\] using the formula: \[\Delta E = \frac{hc}{\lambda}\], where \[h\] is Planck's constant (\[6.626 \times 10^{-34}\] J·s), and \[ c \] is the speed of light (\[3 \times 10^8\] m/s).

Convert Energy Difference to Wavelength

Rearrange the equation \[\Delta E = \frac{hc}{\lambda}\] to solve for \[\lambda\]: \[\lambda = \frac{hc}{\Delta E}\].

Insert Constant Values

Insert the values for \[ h \] and \[c\] in eV\cdot s and m/s respectively, and recognize that 1 eV = 1.602 × 10^(-19) J. The rearranged formula becomes: \[\lambda (A) = \frac{12430}{\Delta E (eV)}\]. This is often approximated as \[\lambda = \frac{12375}{\Delta E}\] Angstroms, due to further constant refinement.

Analyze Choices

Check the problem’s choices against the derived formula. The correct unit for the specific constant provided should be analyzed for conversion to options (1) and (2) against given units (1 Angstrom = 10 Orders \[A = 10^{-10}m = 10^{-8}\text{cm}]\]).

Finalize Your Decision

Recognize (1) fits exact derived with Angstrom, converting to unit Option validation to choices thus match to \[ \frac{12375}{\text{eV}} \].

Key concepts

energy states

Energy states are specific levels of energy that an electron in an atom can have. These levels are quantized, meaning the electron can only exist in certain discrete energy levels, not in between. When an electron absorbs or releases energy, it transitions between these states.

For example, when an electron jumps from a higher energy state to a lower energy state, the difference in energy is released as a photon. This photon's energy is directly related to the wavelength of the emitted light.
The relationship is given by the formula: \[ \Delta E = \frac{hc}{\lambda} \], where:

• \( \Delta E \) is the energy difference between states (in electron volts, eV)
• \( h \) is Planck's constant
• \( c \) is the speed of light
• \( \lambda \) is the wavelength of the emitted light

wavelength determination

Determining the wavelength of light emitted during electronic transitions involves understanding how energy differences translate to specific wavelengths of light.

The formula \( \lambda = \frac{hc}{\Delta E} \) helps us calculate this wavelength. First, we need to know the energy difference \( \Delta E \) between the two states.
By rearranging the formula, we can solve for the wavelength \( \lambda \). Here:

• Insert the values for \( h \) (Planck's constant) and \( c \) (speed of light).
• Make sure \( \Delta E \) is in electron volts (eV).
• Use the approximate constant in Angstroms (\(12375 \)) to simplify the calculation: \( \lambda = \frac{12375}{\Delta E} \text{Å} \).

This allows us to determine the emitted light's wavelength accurately, corresponding to visible, UV, or other spectral lines.

Planck's constant

Planck's constant (\( h \)) is a fundamental constant in quantum mechanics. Its value is approximately \( 6.626 \times 10^{-34} \) J·s. It relates the energy of a photon to its frequency: \( E = hf \).

Planck's constant was foundational for the development of quantum theory. In the context of electronic transitions:

• \( h \) is used to link the energy difference \( \Delta E \) and the wavelength \( \lambda \).
• It helps to convert energy expressed in electron volts into a form usable in our calculations.

Understanding Planck's constant helps grasp concepts of quantum mechanics and the particle nature of light.

speed of light

The speed of light (\( c \)) is another fundamental constant, approximately \( 3 \times 10^8 \) m/s in a vacuum. This constant is critical in physics and helps bridge the gap between energy and wavelength in light emissions.

In the formula \( \Delta E = \frac{hc}{\lambda} \), the speed of light helps express the relationship between the electron's energy states and the light's wavelength.

Here's how it's used in context:

• \( c \) pairs with Planck's constant \( h \) to form the numerator for calculating wavelength.
• When we input \( c \)'s value alongside \( h \), we derive the wavelength for the energy transitions that occur within the atom.

By integrating both \( h \) and \( c \) into our calculations, we accurately determine the characteristics of the emitted photon during electronic transitions.