Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :
![P=\rho gh](https://tex.z-dn.net/?f=P%3D%5Crho%20gh)
Where
is the density of air, ![\rho=1.225\ kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D1.225%5C%20kg%2Fm%5E3)
So,
![P=1.225\times 9.8\times 1\\\\=12\ Pa](https://tex.z-dn.net/?f=P%3D1.225%5Ctimes%209.8%5Ctimes%201%5C%5C%5C%5C%3D12%5C%20Pa)
Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.
Explanation:
Mass of the ball, m = 0.058 kg
Initial speed of the ball, u = 11 m/s
Final speed of the ball, v = -11 m/s (negative as it rebounds)
Time, t = 2.1 s
(a) Let F is the average force exerted on the wall. It is given by :
![F=\dfrac{m(v-u)}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bm%28v-u%29%7D%7Bt%7D)
![F=\dfrac{0.058\times (11-(-11))}{2.1}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B0.058%5Ctimes%20%2811-%28-11%29%29%7D%7B2.1%7D)
F = 0.607 N
(b) Area of wall, ![A=3\ m^2](https://tex.z-dn.net/?f=A%3D3%5C%20m%5E2)
Let P is the average pressure on that area. It is given by :
![P=\dfrac{F}{A}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BF%7D%7BA%7D)
![P=\dfrac{0.607\ N}{3\ m^2}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B0.607%5C%20N%7D%7B3%5C%20m%5E2%7D)
P = 0.202 Pa
Hence, this is the required solution.
Answer:
7500 m/s
Explanation:
We can use the equation velocity of a wave equals wavelength times frequency. Therefore, v = wavelength*f = (25 m)(300 Hz) = m/s7,500