Answer: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N
Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;
![E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%20%29%283%2A10%5E%7B8%7D%29%20%7D%7B270%2A10%5E%7B-9%7D%20%7D%5C%5C%5C%5CE%20%3D%207.362%20%2A%2010%5E%7B-19%7D%20%5C%20J)
The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;
![V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%20%7D%5C%5C%5C%5CV%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%283.677%2A10%5E%7B-19%7D%29%20%7D%7B9.1%2A10%5E%7B-31%7D%20%7D%20%7D%5C%5C%5C%5CV%20%3D%208.99%2A10%5E%7B5%7D%20%20%5C%20m%2Fs)
the shortest de Broglie wavelength for the electrons is given by;
![\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D%5C%5C%20%5C%5C%5Clambda%20%3D%20%5Cfrac%7B6.626%2A10%5E%7B-34%7D%20%7D%7B%289.1%2A10%5E%7B-31%7D%29%28%208.99%2A10%5E%7B5%7D%20%29%7D%5C%5C%5C%5C%5Clambda%20%3D%208.10%2A10%5E%7B-10%7D%20%5C%20m%5C%5C%5C%5C%5Clambda%20%3D%200.81%20%5C%20nm)
Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force ![N= mg cos 22](https://tex.z-dn.net/?f=N%3D%20mg%20cos%2022)
![\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N](https://tex.z-dn.net/?f=%5CRightarrow%20mg%20%3D%5Cfrac%7BN%7D%7Bcos22%7D%3D%5Cfrac%7B538%20N%7D%7B0.927%7D%3D580%20N)
Answer:
red giant stars/ red hyper giants
Explanation:
take stephenson 2-18 for example the star is only 3200k where the sun is around 5000k.