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Umnica [9.8K]
3 years ago
13

What amount of energy is needed for an electron to jump from n = 1 to n = 4?

Physics
1 answer:
liberstina [14]3 years ago
4 0

Answer:

E=2.04\times 10^{-18}\ J

Explanation:

We need to find the energy for an electron to jump from n = 1 to n = 4.

The energy in transition from 1 state to another is given by :

E=\dfrac{-2.18\times 10^{-18}}{n^2}\ J

The difference in energy for n = 1 to n = 4 is:

E=-2.18\times 10^{-18}\times (\dfrac{1}{4^2}-1)\\\\E=2.04\times 10^{-18}\ J

So, the required energy is equal to 2.04\times 10^{-18}\ J.

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A black lift of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35
horsena [70]

Answer:

Explanation:

refractive index of ember = sin of angle of incidence / sin of angle of refraction

= sin 35 / sin24

= .5735 / .4067

= 1.41

This is refractive index of ember with respect to water

refractive index of ember with respect to water

= wμe = μe / μw

μe = wμe x  μw

= 1.33 x 1.41

= 1.87

refractive index of ember with respect to air = 1.87 .

6 0
3 years ago
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

4 0
3 years ago
A metal bar moves through a magnetic field. the induced charges on the bar are
Dmitry [639]
I would say its a positive cgarge
7 0
3 years ago
Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728
Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19  

= 9*1.007 + 10*1.008 = 19.152 u

Actual  mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23  

= 1.38*10^13 J/mol

8 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0
3 years ago
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