Ask question

Ask question

All categories

3 years ago

12

Calculate the work function that requires a 475 nm photon to eject an electron of 1.25 eV.

1 answer:

lapo4ka [179]3 years ago

3 0

The work function has the formula
E = h v

where
E is the energy
h is the constant
v is the frequency

Since the given is wavelength, we use the relationship between frequency and wavelength
v = c / λ
where
c is the speed of light
<span>λ is the wavelength
</span>
So.
E = h c / <span>λ
h = E</span><span>λ / c = 1.25 eV (475x10-9m) / (3x10^8 m/s)
h = 1.9792 x 10^-15

The work function is:
E = 1.9792 x 10-15 v

</span>

You might be interested in

The bright-line spectrum of an element in the gaseous phase is produced as 1. protons move from lower energy states to higher en

lara31 [8.8K]

Answer:

4. Electrons move from higher energy states to lower energy states.

Explanation:

When electrons fall from a higher (excited) energy state to a lower energy state, it loses/gives out energy.

This energy is given out by the emission of photons (quanta of light) by the electron.

7 0

3 years ago

How does the role of blood protect the body

maksim [4K]

Well it transports white blood cells to infected areas.

4 0

3 years ago

A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp

jasenka [17]

Answer:

3.82 ms

Explanation:

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f is the frequency.

In this problem, f = 262 Hz, so the period if this sound wave is

$T=\frac{1}{262 Hz}=3.82\cdot 10^{-3} s=3.82ms$

4 0

3 years ago

A 2 kilogram bowling ball is 2.5 meters off the ground on a post when it falls just before it reaches the ground it is traveling

s2008m [1.1K]

I need the answer too

7 0

3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago