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Yanka [14]
2 years ago
6

Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm

) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?
Physics
1 answer:
kupik [55]2 years ago
7 0

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

\beta=\dfrac{\lambda D}{d}

Put the value into the formula

\beta=\dfrac{0.25\times10^{-9}\times3.3}{0.16\times10^{-3}}

\beta=5.15\times10^{-6}\ m

\beta=5.15\ \mu m

Hence, The spacing is 5.15 μm.

You might be interested in
The wavelength of light is 5000 angstrom. Express it in nm and m.
Ierofanga [76]

Answer:

1 angstrom = 0.1nm

5000 angstrom = 5000/1 × 0.1nm

<h3>= 500nm</h3>

1 \:  angstrom = 1 \times  {10}^{ - 10} m

5000 angstrom = 5000 × 1 × 10^-10

<h3>= 5 × 10^-7 m</h3>

Hope this helps you

7 0
1 year ago
The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an
Troyanec [42]

Answer:

3400 Hz

Explanation:

We know that

1 cm = 0.01 m

L = Length of the human ear canal = 2.5 cm = 0.025 m

V = Speed of sound = 340 ms⁻¹

f = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

f = \frac{(2n - 1)V}{4L}

for first resonant frequency, we have n = 1

Inserting the values

f = \frac{(2(1) - 1) 340}{4(0.025)}

f = \frac{340}{4(0.025)}

f = 3400 Hz

4 0
2 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
1 year ago
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal
iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
1 year ago
An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in
SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

h=-16t^2+v_ot

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

-16t^2+250t=500

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0
2 years ago
Read 2 more answers
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