Nickel has a work function if 5.0eV
A. What is the maximum kinetic energy of photoelectrons knocked out of nickel surface by a 1mW ultraviolet sources at 2000A?
We can use the equation:
Kinetic energy = Energy of incident photon - Work function
The energy of a photon can be calculated using the equation:
Energy of photon = (Planck's constant x speed of light) / wavelength
Where Planck's constant = 6.626 x 10^-34 J s and speed of light = 3 x 10^8 m/s.
Converting the wavelength given to meters:
2000A = 2000 x 10^-10m
Energy of photon = (6.626 x 10^-34 J s x 3 x 10^8 m/s) / (2000 x 10^-10m)
= 9.939 x 10^-19 J
Now we can calculate the kinetic energy of the photoelectrons:
Kinetic energy = (9.939 x 10^-19 J) - (5.0 eV x 1.602 x 10^-19 J/eV)
= 4.939 x 10^-19 J
As the kinetic energy is in joules, it can also be expressed in electronvolts (eV) by dividing the value by the charge of an electron (1.602 x 10^-19 C/electron):
Kinetic energy = (4.939 x 10^-19 J) / (1.602 x 10^-19 C/electron)
= 3.08 eV
Therefore, the maximum kinetic energy of photoelectrons knocked out of nickel surface by a 1mW ultraviolet source at 2000A is 3.08 eV.
hey bot Here are the prices for biweekly lawn mowing services for 13 local companies:
15, 15, 20, 30, 30, 35, 40, 40, 40, 40, 50, 50, 50
Which measure of center—mean or median—best represents the shape of the distribution?
The mean and the median are both measures of center, but which one is more appropriate depends on the shape of the distribution.
In this case, there are a few companies with lower prices (15 and 20), a few with higher prices (50), and several grouped in the middle (between 30-40). This suggests that the distribution might be slightly skewed to the right.
In this situation, the median would be more appropriate than the mean, because the median is less affected by the extreme values of the distribution. In this case, the median is 40, which is a better representation of the center of the distribution than the mean (which is approximately 33.8).
Therefore, the median is the better measure of center for this distribution.
For the given box plot, which measure of variability—range or IQR—best represents the shape of the distribution? (the numbers are 2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5)
The box plot is a graphical representation of the five-number summary, which includes the minimum, maximum, median (Q2), and quartiles (Q1 and Q3) of the distribution.
In this case, the box plot shows that the minimum value is approximately 2, the maximum value is approximately 9.5, and the middle 50% of the data (between Q1 and Q3) is represented by the box. The median is represented by the line inside the box.
The range is calculated by subtracting the minimum value from the maximum value:
Range = Maximum value - Minimum value
In this case, the range is approximately 7.5 (9.5 - 2).
The IQR (interquartile range) is calculated by subtracting the first quartile (Q1) from the third quartile (Q3):
IQR = Q3 - Q1
In this case, the IQR is approximately 4 (Q3 - Q1 = 7 - 3).
The IQR is a better measure of variability than the range for this distribution, because the range is heavily influenced by extreme values. The IQR only considers the middle 50% of the data, and is less affected by outliers. In this case, the IQR represents the spread of the middle 50% of the data, and is a better measure of variability for this distribution.
To calculate the maximum kinetic energy of the photoelectrons, we can use the equation:
Kinetic Energy = Photon Energy - Work Function
First, we need to find the photon energy using the equation:
Photon Energy = Planck's constant * Speed of light / Wavelength
Given:
Planck's constant (h) = 6.626 x 10^-34 J·s
Speed of light (c) = 3 x 10^8 m/s
Wavelength (λ) = 2000 Å (angstroms)
Converting the wavelength to meters:
1 Å = 1 x 10^-10 m
Wavelength (λ) = 2000 x 10^-10 m = 2 x 10^-7 m
Now, we can calculate the photon energy:
Photon Energy = (6.626 x 10^-34 J·s) * (3 x 10^8 m/s) / (2 x 10^-7 m)
Photon Energy ≈ 9.939 x 10^-19 J
Next, we can calculate the maximum kinetic energy of the photoelectrons:
Maximum Kinetic Energy = Photon Energy - Work Function
Given:
Work Function = 5.0 eV
Converting electron volts (eV) to joules (J):
1 eV = 1.602 x 10^-19 J
Work Function = 5.0 eV * (1.602 x 10^-19 J/eV)
Work Function ≈ 8.01 x 10^-19 J
Maximum Kinetic Energy = (9.939 x 10^-19 J) - (8.01 x 10^-19 J)
Maximum Kinetic Energy ≈ 1.929 x 10^-19 J
Therefore, the maximum kinetic energy of the photoelectrons knocked out of the nickel surface by the 1mW ultraviolet source at 2000Å is approximately 1.929 x 10^-19 J.
To find the maximum kinetic energy of photoelectrons knocked out of the nickel surface, we can use the equation:
Kinetic Energy (KE) = Photon Energy - Work Function
Step 1: Convert the power of the ultraviolet source from milliwatts (mW) to watts (W).
Given:
Power of the ultraviolet source = 1 mW = 1 × 10^(-3) W
Step 2: Calculate the energy of one photon using the formula:
Energy of one photon (E) = Planck's constant (h) × speed of light (c) / wavelength
Given:
Wavelength (λ) = 2000 Å (angstroms)
Planck's constant (h) = 6.626 × 10^(-34) J·s
Speed of light (c) = 3 × 10^8 m/s
First, we need to convert the wavelength from angstroms to meters:
2000 Å = 2000 × 10^(-10) m = 2 × 10^(-7) m
Now we can calculate the energy of one photon:
E = (6.626 × 10^(-34) J·s) × (3 × 10^8 m/s) / (2 × 10^(-7) m)
Step 3: Calculate the maximum kinetic energy:
Maximum Kinetic Energy (KE) = Energy of one photon - Work Function
Given:
Work Function of nickel (φ) = 5.0 eV
We need to convert the work function from electron volts (eV) to joules (J):
1 eV = 1.602 × 10^(-19) J
5.0 eV = 5.0 × 1.602 × 10^(-19) J = 8.01 × 10^(-19) J
Now we can calculate the maximum kinetic energy:
KE = E - φ
Finally, we substitute the values to calculate the maximum kinetic energy.