The differential cross section for scattering 6.5-mev alpha particles at 120° off a silver nucleus is about 0.5 barns/sr. if a total of 10 10 alphas impinge on a silver foil of thickness 1 itm and if we detect the scattered particles using a counter of area 0.1 mm 2 at 120° and 1 cm from the target, about how many scattered alphas should we expect to count? (silver has a specific gravity of 10.5, and atomic mass of 108.) 14.12 ***

Nsc=30

Explanation:

The solid angle subtended by the counter is

d=0.1mm2/(1cm)2=(10)3 sr

d=(10)10x10.5gcm3x10-4cm/108x1.6610x10-24gx0.510-24cm2x10-3=30

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.