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spayn [35]
4 years ago
12

There is an increasing demand for berries in Paul’s town. He wants to cultivate berries to earn profits. To do so, he wants to c

heck whether the soil in his farm is suitable for cultivating berries. Which characteristic of soil will most accurately check the suitability of land for growing berries?
A) soil profile
B) soil texture
C ) soil horizon
D) soil moisture
Physics
1 answer:
iren [92.7K]4 years ago
8 0

Answer:

Option (B)

Explanation:

Soil texture is defined as the amount of soil that is present and the type of materials that comprises the soil such as sand, silt and clay particles, of which the sands are the coarser particles and the clay is the finer particles.

This soil texture enables us to determine the properties such as moisture content,  porosity and permeability and the number of nutrients that it can provide for the growth and expansion of plants.

Thus, from the above-given condition, the soil texture property helps in checking the suitability of the land for growing berries.

Thus, the correct answer is option (B).

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3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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Answer:

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Its SI unit is newton(n).

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