Ask question

Ask question

All categories

3 years ago

9

What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=

2.93 eV. Tries 0/20 In this experiment, what is the minimal potential needed to fully stop the electrons if the wavelength of the incident light is 321 nm?

1 answer:

zalisa [80]3 years ago

7 0

Answer:

4.24nm

0.385eV

Explanation:

Maximum wavelength (λmax) :

λmax = ( hc) /Φ

h = plancks constant = 6.63 * 10^-34

c = speed of light = 3*10^8

1ev = 1.6 * 10^-19

Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19

λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19

λmax = 19.89 * 10^-26 / 4.688*10^-19

λmax = 4.242 * 10^-7 m

λmax= 4.24nm

B.)

E = hc / eλ eV

λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9

E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)

E = 19.89 * 10^-26 / 600 * 10^-28

E = 0.03315 * 10^-26 + 28

E = 0.03315 * 10^2

E = 3.315 eV

Stopping potential : (3.315 eV - 2.93eV) = 0.385eV

You might be interested in

Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter.

padilas [110]

Answer:

$20.32^{\circ}$ and $44.08^{\circ}$

$12.56^{\circ}$ and $25.77^{\circ}$

Explanation:

$\lambda$ = Wavelength

$\theta$ = Angle

m = Order

Distance between grating is given by

$d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}$

$\lambda=656\ \text{nm}$

We have the relation

$d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}$

The first and second order angular deflection is $20.32^{\circ}$ and $44.08^{\circ}$

$\lambda=410\ \text{nm}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}$

The first and second order angular deflection is $12.56^{\circ}$ and $25.77^{\circ}$ .

4 0

3 years ago

Calculate the correct fuse that should be used for a 230V,1KW electric hair dryer.

zzz [600]

Answer: 4 A

Explanation:

Given

Voltage $V=230\ V$

Power $P=1\ kW$

Power is given by $P=VI\\$

Insert the values

$\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A$

The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.

6 0

3 years ago

The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?

bezimeni [28]

Answer:

660V

Explanation:

V=IR

V=11×60

=660V

hope this helps

please like and Mark as brainliest

8 0

4 years ago

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

slavikrds [6]

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is $p=7.2\times 10^{-3}\ kg-m/s$

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, $p=7.2\times 10^{-3}\times 10^3\ g-m/s$

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

6 0

4 years ago

You hear the sound of the firing of a distant cannon 6.9 s after seeing the flash. how far are you from the cannon? (assume 343

EastWind [94]

Let's call d the distance of the cannon from us. We see the flash first, then after 6.9 s we hear the sound. The flash arrives to us traveling at speed of light, which is $c=3 \cdot 10^8 m/s$ , while the sound travels at speed $v=343 m/s$ . Since the magnitude of the speed of light is much much larger than the speed of sound, we can assume that the flash arrives at our location instantaneously after the firing.
Then we can say that the sound wave covers a distance d traveling at speed v in a time of t=6.9 s. Since it is a uniformly linear motion, the distance covered by the sound wave is
$d=vt=(343 m/s)(6.9 s)=2367 m$
So, this is the distance of the cannon from us.

6 0

3 years ago