atomic structure chapter need answers and solutions urgent 6ans:1113×10^21 7ans:6.41×10^24 8ans:6.92×10^-19 joules
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Que : Find the number of quanta .....
Solution :
Energy of quanta,
E = hν
= 6.626 × 10⁻³⁴ Js × 4.75 × 10¹³ s⁻¹
= 31.5 × 10 ⁻²¹ J
It is given that energy required to melt 1 g of ice = 350 J
So, energy required to melt 100 g of ice = 35000 J
No. of quanta in 35000 J of energy = 35000 J / 31.5 × 10 ⁻²¹ J = 1.12 × 10 ²⁴ (Ans.)
Solution :
Energy of quanta,
E = hν
= 6.626 × 10⁻³⁴ Js × 4.75 × 10¹³ s⁻¹
= 31.5 × 10 ⁻²¹ J
It is given that energy required to melt 1 g of ice = 350 J
So, energy required to melt 100 g of ice = 35000 J
No. of quanta in 35000 J of energy = 35000 J / 31.5 × 10 ⁻²¹ J = 1.12 × 10 ²⁴ (Ans.)
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6. Energy in one photon or quantum of radiation = h v
E = 6.623 * 10⁻³⁴ * 4.75 * 10¹³ Joules.
energy to melt 100 gm of ice = 35, 000 J
Number of quanta =35,000/ E.
7. Energy emitted in 10 hrs = E = 60 W * 10 * 3600 sec
energy in one quantum /photon = Ep
= hc/λ = 6.623 * 10⁻³⁴ * 3 * 10⁸ / (5893 * 10⁻¹⁰) J
number of photons = E / Ep
8. If the electrons absorb all the energy in the photons of the radiation, the energy that remains after overcoming the attractive forces of atoms on the surface will be the K.E.
K.E. <= h c / λ - 7.52 * 10⁻¹⁹ J
= 6.623 * 10⁻³⁴ * 3 * 10⁸ / 360 * 10⁻¹⁰ - 7.52 * 10⁻¹⁹ J
E = 6.623 * 10⁻³⁴ * 4.75 * 10¹³ Joules.
energy to melt 100 gm of ice = 35, 000 J
Number of quanta =35,000/ E.
7. Energy emitted in 10 hrs = E = 60 W * 10 * 3600 sec
energy in one quantum /photon = Ep
= hc/λ = 6.623 * 10⁻³⁴ * 3 * 10⁸ / (5893 * 10⁻¹⁰) J
number of photons = E / Ep
8. If the electrons absorb all the energy in the photons of the radiation, the energy that remains after overcoming the attractive forces of atoms on the surface will be the K.E.
K.E. <= h c / λ - 7.52 * 10⁻¹⁹ J
= 6.623 * 10⁻³⁴ * 3 * 10⁸ / 360 * 10⁻¹⁰ - 7.52 * 10⁻¹⁹ J
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