1.31 It is now established that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (–1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

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alok kumar singh | Contributor-Level 10

3 months ago

1.31 A proton has 3 quarks. Let there be n ‘up quarks’, then number of ‘down quarks’ = 3-n

Charge due to n ‘up quarks’ = $(\frac{2}{3}$ e) n

Charge due to (3-n) ‘down quarks’ = $-$ $\frac{1}{3} e) (3 - n)$

Total charge on a proton = +e = $(\frac{2}{3}$ e) n + ( - $\frac{1}{3} e) (3 - n)$

e = $\frac{2 n e}{3}$ + $\frac{n e}{3}$ -e

2e = $\frac{3 n e}{3}$

n = 2

Number of ‘up quark’ = 2 and number of ‘down quark’ = 1. Therefore a proton can be represented as ‘uud’.

A neutron has 3 quarks. Let there be n ‘up quark’ in a neutron and (3-n) ‘down quark’

Charge due to n ‘up quark’ = + $(\frac{2}{3}$ e)n

Charge due to (3-n) ‘down quark’

...more

1.31 A proton has 3 quarks. Let there be n ‘up quarks’, then number of ‘down quarks’ = 3-n

Charge due to n ‘up quarks’ = $(\frac{2}{3}$ e) n

Charge due to (3-n) ‘down quarks’ = $-$ $\frac{1}{3} e) (3 - n)$

Total charge on a proton = +e = $(\frac{2}{3}$ e) n + ( - $\frac{1}{3} e) (3 - n)$

e = $\frac{2 n e}{3}$ + $\frac{n e}{3}$ -e

2e = $\frac{3 n e}{3}$

n = 2

Number of ‘up quark’ = 2 and number of ‘down quark’ = 1. Therefore a proton can be represented as ‘uud’.

A neutron has 3 quarks. Let there be n ‘up quark’ in a neutron and (3-n) ‘down quark’

Charge due to n ‘up quark’ = + $(\frac{2}{3}$ e)n

Charge due to (3-n) ‘down quark’ =
- ( $\frac{1}{3} e) (3 - n)$

Since total charge of a neutron is zero, we get

+ $(\frac{2}{3}$ e)n -( $\frac{1}{3} e) (3 - n)$ = 0

$(\frac{2}{3}$ e)n = ( $\frac{1}{3} e) (3 - n)$

$\frac{2 e n}{3} = e - \frac{e n}{3}$

en = e or n = 1

Hence number of ‘up quark’ in neutron = 1

Number of ‘down quark’ in neutron = 3-1 = 2

Therefore a neutron can be represented as ‘udd’.

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1.31 A proton has 3 quarks. Let there be n ‘up quarks’, then number of ‘down quarks’ = 3-nCharge due to n ‘up quarks’ = <math><mo>(</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> e) nCharge due to (3-n) ‘down quarks’ =<math><mo>-</mo></math><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>e</mi><mo>)</mo><mo>(</mo><mn>3</mn><mo>-</mo><mi>n</mi><mo>)</mo></math> Total charge on a proton = +e = <math><mo>(</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> e) n +  ( -<math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>e</mi><mo>)</mo><mo>(</mo><mn>3</mn><mo>-</mo><mi>n</mi><mo>)</mo></math>  e =  <math><mfrac><mrow><mrow><mn>2</mn><mi>n</mi><mi>e</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> +<math><mfrac><mrow><mrow><mi>n</mi><mi>e</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math>-e2e = <math><mfrac><mrow><mrow><mn>3</mn><mi>n</mi><mi>e</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math>n = 2Number of ‘up quark’ = 2 and number of ‘down quark’ = 1. Therefore a proton can be represented as ‘uud’.A neutron has 3 quarks. Let there be n ‘up quark’ in a neutron and (3-n) ‘down quark’Charge due to n ‘up quark’ = +<math><mo>(</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> e)nCharge due to (3-n) ‘down quark’ =  - ( <math><mo><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>e</mi><mo>)</mo><mo>(</mo><mn>3</mn><mo>-</mo><mi>n</mi><mo>)</mo></math></mo></math>Since total charge of a neutron is zero, we get+ <math><mo>(</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math>e)n -(<math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>e</mi><mo>)</mo><mo>(</mo><mn>3</mn><mo>-</mo><mi>n</mi><mo>)</mo></math> = 0 <math><mo>(</mo><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> e)n = ( <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>e</mi><mo>)</mo><mo>(</mo><mn>3</mn><mo>-</mo><mi>n</mi><mo>)</mo></math><math><mfrac><mrow><mrow><mn><math><mfrac><mrow><mrow><mn>2</mn><mi>e</mi><mi>n</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mo>=</mo><mi>e</mi><mo>-</mo><mfrac><mrow><mrow><mi>e</mi><mi>n</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></mn></mrow></mrow></mfrac></math>en = e or n = 1Hence number of ‘up quark’ in neutron = 1Number of ‘down quark’ in neutron = 3-1 = 2Therefore a neutron can be represented as ‘udd’.

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