Answer :
Let's break the problem into two parts and solve them step-by-step.
### Part 1: Calculate t
Given:
- The energy of a radiation of wavelength [tex]\( 8000 \text{ \AA} \)[/tex] is [tex]\( E_1 \)[/tex].
- The energy of a radiation of wavelength [tex]\( 16000 \text{ \AA} \)[/tex] is [tex]\( E_2 \)[/tex].
- The relationship between energies [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex] is [tex]\( E_1 = t \times E_2 \)[/tex].
First, you need to convert the wavelengths from Angstroms ([tex]\( \text{\AA} \)[/tex]) to meters (m):
- [tex]\( 8000 \text{ \AA} = 8000 \times 10^{-10} \text{ m} = 8 \times 10^{-7} \text{ m} \)[/tex]
- [tex]\( 16000 \text{ \AA} = 16000 \times 10^{-10} \text{ m} = 1.6 \times 10^{-6} \text{ m} \)[/tex]
The energy of a photon is given by [tex]\( E = \frac{hc}{\lambda} \)[/tex], where [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)[/tex]) and [tex]\( c \)[/tex] is the speed of light ([tex]\( 3 \times 10^8 \text{ m/s} \)[/tex]).
Now, calculate [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex]:
[tex]\( E_1 \)[/tex]:
[tex]\[ E_1 = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3 \times 10^8 \text{ m/s})}{8 \times 10^{-7} \text{ m}} = 2.48475 \times 10^{-19} \text{ J} \][/tex]
[tex]\( E_2 \)[/tex]:
[tex]\[ E_2 = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3 \times 10^8 \text{ m/s})}{1.6 \times 10^{-6} \text{ m}} = 1.242375 \times 10^{-19} \text{ J} \][/tex]
To find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{E_1}{E_2} = \frac{2.48475 \times 10^{-19} \text{ J}}{1.242375 \times 10^{-19} \text{ J}} = 2.0 \][/tex]
So, the value of [tex]\( t \)[/tex] is [tex]\( 2.0 \)[/tex].
### Part 2: Calculate the frequency of a photon with energy 2 electron volts (eV)
Given:
- Energy, [tex]\( E \)[/tex], of the photon is [tex]\( 2 \text{ eV} \)[/tex].
- 1 electron volt (\text{ eV}) equals [tex]\( 1.602 \times 10^{-19} \text{ J} \)[/tex].
Convert the energy to joules:
[tex]\[ E = 2 \text{ eV} = 2 \times 1.602 \times 10^{-19} \text{ J} = 3.204 \times 10^{-19} \text{ J} \][/tex]
The frequency [tex]\( f \)[/tex] of the photon is given by [tex]\( f = \frac{E}{h} \)[/tex], where [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)[/tex]).
Calculate the frequency [tex]\( f \)[/tex]:
[tex]\[ f = \frac{3.204 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J} \cdot \text{s}} = 4.8354965288258375 \times 10^{14} \text{ Hz} \][/tex]
So, the frequency of a photon having an energy of 2 electron volts is approximately [tex]\( 4.835 \times 10^{14} \text{ Hz} \)[/tex].
### Part 1: Calculate t
Given:
- The energy of a radiation of wavelength [tex]\( 8000 \text{ \AA} \)[/tex] is [tex]\( E_1 \)[/tex].
- The energy of a radiation of wavelength [tex]\( 16000 \text{ \AA} \)[/tex] is [tex]\( E_2 \)[/tex].
- The relationship between energies [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex] is [tex]\( E_1 = t \times E_2 \)[/tex].
First, you need to convert the wavelengths from Angstroms ([tex]\( \text{\AA} \)[/tex]) to meters (m):
- [tex]\( 8000 \text{ \AA} = 8000 \times 10^{-10} \text{ m} = 8 \times 10^{-7} \text{ m} \)[/tex]
- [tex]\( 16000 \text{ \AA} = 16000 \times 10^{-10} \text{ m} = 1.6 \times 10^{-6} \text{ m} \)[/tex]
The energy of a photon is given by [tex]\( E = \frac{hc}{\lambda} \)[/tex], where [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)[/tex]) and [tex]\( c \)[/tex] is the speed of light ([tex]\( 3 \times 10^8 \text{ m/s} \)[/tex]).
Now, calculate [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex]:
[tex]\( E_1 \)[/tex]:
[tex]\[ E_1 = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3 \times 10^8 \text{ m/s})}{8 \times 10^{-7} \text{ m}} = 2.48475 \times 10^{-19} \text{ J} \][/tex]
[tex]\( E_2 \)[/tex]:
[tex]\[ E_2 = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3 \times 10^8 \text{ m/s})}{1.6 \times 10^{-6} \text{ m}} = 1.242375 \times 10^{-19} \text{ J} \][/tex]
To find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{E_1}{E_2} = \frac{2.48475 \times 10^{-19} \text{ J}}{1.242375 \times 10^{-19} \text{ J}} = 2.0 \][/tex]
So, the value of [tex]\( t \)[/tex] is [tex]\( 2.0 \)[/tex].
### Part 2: Calculate the frequency of a photon with energy 2 electron volts (eV)
Given:
- Energy, [tex]\( E \)[/tex], of the photon is [tex]\( 2 \text{ eV} \)[/tex].
- 1 electron volt (\text{ eV}) equals [tex]\( 1.602 \times 10^{-19} \text{ J} \)[/tex].
Convert the energy to joules:
[tex]\[ E = 2 \text{ eV} = 2 \times 1.602 \times 10^{-19} \text{ J} = 3.204 \times 10^{-19} \text{ J} \][/tex]
The frequency [tex]\( f \)[/tex] of the photon is given by [tex]\( f = \frac{E}{h} \)[/tex], where [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)[/tex]).
Calculate the frequency [tex]\( f \)[/tex]:
[tex]\[ f = \frac{3.204 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J} \cdot \text{s}} = 4.8354965288258375 \times 10^{14} \text{ Hz} \][/tex]
So, the frequency of a photon having an energy of 2 electron volts is approximately [tex]\( 4.835 \times 10^{14} \text{ Hz} \)[/tex].
