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3 years ago

13

Which electron transitions in the Bohr hydrogen atom will produce photons with the shortest wavelengths?

1 answer:

Oliga [24]3 years ago

8 0

Those that take the electron from the highest orbital to the lowest (n=1) at this point the energy is -13.6eV. The wavelength is given by c/f. So a large energy (E=hf) gives a short wavelength. Falling from a farther distance out down to the nucleus releases the most energy

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A 12 kg box is at rest on your kitchen counter, which your cat is pawing at with a horizontal force of 40 N. If the coefficient

murzikaleks [220]

Answer; I think it's False.

8 0

3 years ago

An object is thrown off a cliff with a horizontal speed of 10 m/sec. After 3 seconds the object hits the ground. Find the height

Degger [83]

Answer:

223

Explanation:

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4 0

3 years ago

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Nastasia [14]

Answer:

750 nm

Explanation:

$d$ = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

$D$ = screen distance = 4.8 m

$y$ = position of first bright fringe = $\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m$

$n$ = order = 1

Position of first bright fringe is given as

$y = \frac{nD\lambda }{d}$

$0.002 = \frac{(1)(4.8)\lambda }{0.0018}$

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0

3 years ago

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

liberstina [14]

Answer:

2C

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

5 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago