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š´ÊŠ 4 ‹»—Á„ȝ˜´ª°¥nµŠÄœ¦³—´®¨´„ nªœ˜Îµ®œ· ¸Ÿ·—ž„˜·¡š¸É ‹»—„µ¦‡´— œµ—¨³˜¦ª‹°‡» –£µ¡ ¨³˜Îµ®œ· ÁœºÊ°´ ¤Ÿ´
¡š´Ê Š 4 ‹»—Á„ȝ˜´ª°¥nµŠ¤¸‡ªµ¤Îµ‡´ ¦³—´š¸É µ¤µ¦™‡ª‡» ¤Å—o 宦´˜´ª°¥nµŠ‹µ„Á¦º °¨µ„‡¼n¡˜Îµ®œ· „ž¦„š¸É‹»—„µ¦
‡´— œµ—¨³˜¦ª‹°‡»–£µ¡Äœ¦³—´ ‡ª‡»¤Å—o –³š¸É˜´ª°¥nµŠ‹µ„Á¦º °Å—®r¡˜Îµ®œ· „ž¦„š¸É‹»—„µ¦‡´ — œµ—¨³
˜¦ª‹°‡»–£µ¡Â¨³š¸É‹»—„µ¦ž{~œž¨µ®¤¹„ Ĝ¦³—´‡ª‡»¤Å—o ¨³¦³—´¦°Š nªœ˜Îµ®œ· ·ÉŠÂž¨„ž¨°¤¡š¸É‹»—„µ¦‡´—
œµ—¨³˜¦ª‹°‡» –£µ¡¤¸‡ªµ¤Îµ‡´ ¦³—´š¸É µ¤µ¦™‡ª‡»¤Å—o
˜µ¦µŠš¸É 4 ¨Îµ—´‡ªµ¤Îµ‡´ °ŠŸ¨„¦³š °Š˜Î µ®œ·‹µ„Á¦º°š»„œ· —
œ·—˜Îµ®œ· ‹»—Á„ȝ˜´ª°¥nµŠ ‡³Âœœ
(¦o°¥¨³ °Šž¦· ¤µ–˜Îµ®œ·) O x S ‡ªµ¤Î µ‡´
Á¦º°¨µ„Á—¸É ¥ª
1.„ž¦„ 2 (30.05) 5 x 4 = 20 Ma (®¨´„)
2.„ž¦„ 3 (15.02) 5 x 4 = 20 Ma (®¨´„)
3.„ž¦„ 4 (13.87) 5 x 4 = 20 Ma (®¨´„)
4.„ž¦„ 1 (12.01) 5 x 4 = 20 Ma (®¨´„)
5.¸Ÿ·—ž„˜· 3 (7.83) 3 x 3 = 9 Co (‡ª‡»¤Å—o )
6.ÁœºÊ°´¤Ÿ´ 1 (5.87) 2 x 3 = 6 Co (‡ª‡»¤Å—o )
7.ÁœºÊ°´¤Ÿ´ 3 (5.83) 2 x 3 = 6 Co (‡ª‡»¤Å—o )
8.ÁœºÊ°´¤Ÿ´ 4 (4.87) 2 x 3 = 6 Co (‡ª‡»¤Å—o )
9.ÁœºÊ°´¤Ÿ´ 2 (3.80) 2 x 3 = 6 Co (‡ª‡»¤Å—o )
Á¦º°¨µ„‡¼n
1.„ž¦„ 3 (3.27) 2 x 4 = 8 Co (‡ª‡»¤Å—o )
Á¦º°Å—®r
1.„ž¦„ 4 (8.60) 3 x 4 = 12 Mi (¦°Š)
2.„ž¦„ 3 (3.30) 2 x 4 = 8 Co (‡ª‡»¤Å—o )
3.·ÉŠÂž¨„ž¨°¤ 3 (8.37) 3 x 2 = 6 Co (‡ª‡»¤Å—o )
®¤µ¥Á®˜»: 1 = ¦´ª´˜™»—·2 = —°Š¦°Ÿ¨·˜ 3 = ‡´— œµ—¨³˜¦ª‹°‡»–£µ¡4 =ž{~œž¨µ®¤¹„ O= ð„µ S = ‡ªµ¤¦»œÂ¦Š
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