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.
Answer:
They have a dual wave-particle nature.
Explanation:
Electromagnetic waves consist of periodic oscillations of electric and magnetic field in a plane perpendicular to the direction of motion of the wave (in fact, they are also classified as transverse waves).
Electromagnetic waves have a wave nature, however they also have particle nature - in fact, it has been proved in some experiment (e.g. photoelectric effect) that in some conditions they act as packets of particles - called photons. Therefore, the option
They have a dual wave-particle nature.
is correct.
Other options are wrong because:
They are all invisible. --> False because visible light (which is part of the electromagnetic spectrum, so they are electromagnetic waves) is visible
They can only travel without a medium. --> False because they can also travel in a vacuum
They are slower than sound waves. --> False because they travel much faster (they travel at the speed of light in a vacuum,
, while sound travels at 343 m/s in air, for instance)
To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.
A) Conservation of Energy,


Here,
m = Mass
v = Velocity
k = Spring constant
A = Amplitude
Rearranging to find the Amplitude we have,

Replacing,


(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.
The Period is defined as

Replacing,


Now the velocity is described as,


We have all the values, then replacing,


Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm