# Maple TA 24_2

## Maple TA 24_2

Hey guys, here's a head start on Maple TA 24_2

I'm still trying to figure out # 1

2. In a Millikan type experiment a charge of -1.625 x10-18C is measured for the oildrop. How many electrons carry this charge (sheet 19) ? Do not round to get an integer number. Indicate with a positive (negative) sign whether this number corresponds to the number of excess electrons (the total number of electrons) contained in the drop.

=(-1.625E-18)/(1.6E-19)

= -10.15625 (NEGATIVE)

I'm still trying to figure out # 1

2. In a Millikan type experiment a charge of -1.625 x10-18C is measured for the oildrop. How many electrons carry this charge (sheet 19) ? Do not round to get an integer number. Indicate with a positive (negative) sign whether this number corresponds to the number of excess electrons (the total number of electrons) contained in the drop.

=(-1.625E-18)/(1.6E-19)

= -10.15625 (NEGATIVE)

**User X**- Guest

## Maple TA 24_2

3. You start out with 162.7 radioactive nuclei. After 1.223 minutes 87.95 are left over. What is the average activity (sheet 22) ?

= (162.7-87.95)/(1.223*60)

= 1.018669937(POSITIVE)

= (162.7-87.95)/(1.223*60)

= 1.018669937(POSITIVE)

**User X**- Guest

## Maple TA 24_2

4. A sample of radioactive nuclei with a decay constant of 1.281 hour-1 has an activity of 74.41 decays per second. How many radioactive nuclei does the sample contain (sheet 22) ? Indicate with a positive (negative) sign whether the number of radioactive nuclei increases (decreases) with time.

= (74.41)/(1.281/3600)

= -209114.7541 (NEGATIVE)

the reason you divide the 1.281hrs^-1 is in order to get it in seconds

= (74.41)/(1.281/3600)

= -209114.7541 (NEGATIVE)

the reason you divide the 1.281hrs^-1 is in order to get it in seconds

**User X**- Guest

## Maple TA 24_2

for #2, i didn't realize it came up as a smiley face....so here's the full equation:

2. In a Millikan type experiment a charge of -1.625 x10-18C is measured for the oildrop. How many electrons carry this charge (sheet 19) ? Do not round to get an integer number. Indicate with a positive (negative) sign whether this number corresponds to the number of excess electrons (the total number of electrons) contained in the drop.

=(-1.625E-18 ) / (1.6E-19)

= -10.15625 (NEGATIVE)

2. In a Millikan type experiment a charge of -1.625 x10-18C is measured for the oildrop. How many electrons carry this charge (sheet 19) ? Do not round to get an integer number. Indicate with a positive (negative) sign whether this number corresponds to the number of excess electrons (the total number of electrons) contained in the drop.

=(-1.625E-18 ) / (1.6E-19)

= -10.15625 (NEGATIVE)

**User X**- Guest

## Re: Maple TA 24_2

Question 1:

(Mass x Gravity)/(Charge)

Answer is negative. Distance is not used for this problem.

(Mass x Gravity)/(Charge)

Answer is negative. Distance is not used for this problem.

**helping**- Guest

## number 2 sign

hey, thanks for the solution for number 2, i confirmed it with mine, but for some reason my mapleta accepted the answer as postive instead o.o

**:)**- Guest

## Re: Maple TA 24_2

#2 is definitely positive, the question asks for it to be positive if what you calculated were excess electrons, and on the CD the page has the word "excess" in bold right on the page with the formula.

**Andrew**- Guest

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