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2 years ago

What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?

Physics

1 answer:

Sindrei [870]2 years ago

7 0

Answer:

$9.12\cdot 10^{-8} m$

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

$E=13.6 eV= 2.18\cdot 10^{-18}J$

The energy of the photon is related to the wavelength by

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

Solving the formula for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m$

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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should

liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

$v_x = 14 cos30 = 12.12 m/s$

$v_y = 14 sin30 = 7 m/s$

vertical distance between them

$\delta y = 4 m$

now we will use kinematics in order to find the time taken by the ball to reach at Allison

$\delat y = v_y *t + \frac{1}{2} at^2$

here acceleration is due to gravity

$a = 9.8 m/s^2$

now we will have

$4 = 7 * t + \frac{1}{2}*9.8 * t^2$

now solving above quadratic equation we have

$t = 0.44 s$

now in order to find the horizontal distance where ball will fall is given as

$d = v_x * t$

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

$d = 12.12 * 0.44 = 5.33 m$

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0

2 years ago

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

$a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev$

$d= 155 rev = 155(2\pi ) = 310\pi rad$

$a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi ) = 4.0\pi \frac{rev}{s^{2} }$

$d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449$

$w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}$

Now the angular velocity is the blade speed so:

$V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}$

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

$t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s$

Now is the same time the piece travel horizontally

$L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m$

blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS

6 0

2 years ago

A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?

natita [175]

Answer:

2.08 kg

Explanation:

Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

$F=ma$

where F is the force applied, m is the mass of the object and a its acceleration.

In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

$m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg$

6 0

2 years ago

In the Roman soldier model for refraction, what happens to the first soldier who hits the muddy stream?.....HELP PLZ.... A. They

SpyIntel [72]

Answer:

uKkskdjod 7q and the rays are the best in all the ways ❤ ♥

7 0

2 years ago